Chapter 1.4

Miscellaneous Problems

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Problem 1.140

Suppose that f(x)f(x) is a real function of the real variable xx. If the functions ∣f′(x)∣,∣f′′(x)∣,…\left|f'(x)\right|, \left|f''(x)\right|, \ldots are steadily and strictly decreasing in the interval [0,x][0,x], x>0x>0, then f(x)f(x) is enveloped in the strict sense by its Maclaurin series.

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Problem 1.141

The functions

e−x;log⁡(1+x);(1+x)−p,p>0; e^{-x}; \quad \log (1+x); \quad (1+x)^{-p}, \qquad p>0 ;

are strictly enveloped by their Maclaurin series for x>0x>0.

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Problem 1.142

(continued). Prove the same for the functions

cos⁡x,sin⁡x. \cos x, \quad \sin x .
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Remark.
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Obviously only the non-vanishing terms of the Maclaurin series are to be considered. E.g.the nn-th partial sum of the Maclaurin series for cos⁡x\cos x is

1−x22!+x44!−⋯+(−1)nx2n(2n)!,n=0,1,2,… 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^n \frac{x^{2n}}{(2n)!} , \qquad n=0,1,2,\ldots
Problem 1.143

(continued). Prove the same for the functions

arctan⁡x,J0(x)=1−11!1!(x2)2+12!2!(x2)4−⋯[problems 141, 142]. \arctan x, \qquad J_0(x) = 1 - \frac{1}{1!1!}\left(\frac{x}{2}\right)^2 + \frac{1}{2!2!}\left(\frac{x}{2}\right)^4 - \cdots \qquad \text{[problems 141, 142].}
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Problem 1.144

Suppose that the terms of the series a0+a1+a2+⋯a_0+a_1+a_2+\cdots are alternately positive and negative and that there exists a number AA such that

A−(a0+a1+a2+⋯+an) A - (a_0+a_1+a_2+\cdots +a_n)

assumes always the same sign as the next term, an+1a_{n+1}. Then the series envelops AA in the strict sense.

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Problem 1.145

If the series a0+a1+a2+⋯a_0+a_1+a_2+\cdots, ana_n real, n=0,1,2,…n=0,1,2,\ldots, envelops the real number AA and if in addition ∣a1∣>∣a2∣>∣a3∣>⋯\left|a_1\right|>\left|a_2\right|>\left|a_3\right|>\cdots then the terms a1,a2,a3,…a_1,a_2,a_3,\ldots have alternating signs and AA is enveloped in the strict sense.

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Problem 1.146

Let the function f(x)f(x) assume real values for real xx, x>R>0x>R>0. If f(x)f(x) is enveloped for x>Rx>R by the real series a0+a1x+a2x2+a3x3+⋯a_0+\dfrac {a_1}{x}+\dfrac {a_2}{x^2}+\dfrac {a_3}{x^3}+\cdots then the numbers a1,a2,a3,…a_1,a_2,a_3,\ldots have alternating signs and the series is strictly enveloping.

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Problem 1.147

Suppose that the real valued function f(t)f(t) is infinitely often differentiable for t≥0t \geq 0 and that all its derivatives f(n)(t)f^{(n)}(t) (n=0,1,2,…n=0,1,2,\ldots) have decreasing absolute values and converge to 00 for t→∞t \rightarrow \infty. Then the integral

∫0∞f(t)cos⁡xt dt \int _0^{\infty } f(t) \cos xt \, dt

is, for real xx, strictly enveloped by the series

−f′(0)x2+f′′′(0)x4−fV(0)x6+fVII(0)x8−⋯ . -\frac{f'(0)}{x^2} + \frac{f'''(0)}{x^4} - \frac{f^{\mathrm{V}}(0)}{x^6} + \frac{f^{\mathrm{VII}}(0)}{x^8} - \cdots .
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Remark.
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Example: f(t)=e−tf(t) = e^{-t}.

Problem 1.148

The number 23\dfrac {2}{3} is enveloped by the series

34+14−38−18+316+116−332−132+⋯ , \frac{3}{4} + \frac{1}{4} - \frac{3}{8} - \frac{1}{8} + \frac{3}{16} + \frac{1}{16} - \frac{3}{32} - \frac{1}{32} + \cdots ,

but not in the strict sense.

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Problem 1.149

Plot the first seven terms of the series

ei=1+i1!−12!−i3!+14!+i5!−16!−⋯ e^i = 1 + \frac{i}{1!} - \frac{1}{2!} - \frac{i}{3!} + \frac{1}{4!} + \frac{i}{5!} - \frac{1}{6!} - \cdots

successively as complex numbers and compute so the value of eie^i to three decimals.

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Problem 1.150

Let H\mathfrak {H} denote a ray with origin z=0z=0. Assume that along H\mathfrak {H} all the derivatives of the function f(z)f(z) attain the maximum of their absolute values at the origin and only there; i.e.

∣f(n)(0)∣>∣f(n)(z)∣ \left|f^{(n)}(0)\right| > \left|f^{(n)}(z)\right|

whenever zz is on H\mathfrak {H} and ∣z∣>0\left|z\right|>0. Then:

  1. The function f(t)f(t) is enveloped for every zz on H\mathfrak {H} by the Maclaurin series

    f(0)+f′(0)1!z+f′′(0)2!z2+⋯[problem 140]. f(0) + \frac{f'(0)}{1!} z + \frac{f''(0)}{2!} z^2 + \cdots \qquad \text{[problem 140].}
  2. the function F(z)=∫0∞e−tf(tz)dtF(z) = \displaystyle \int_0^{\infty } e^{-t} f\left(\frac{t}{z}\right) dt is for every zz on H‾\overline{\mathfrak {H}} enveloped by the series

    f(0)+f′(0)z+f′′(0)z2+f′′′(0)z3+⋯[problem 147]; f(0) + \frac{f'(0)}{z} + \frac{f''(0)}{z^2} + \frac{f'''(0)}{z^3} + \cdots \qquad \text{[problem 147];}

    H‾\overline{\mathfrak {H}} is the mirror image of H\mathfrak {H} with respect to the xx-axis; the integral is taken along the positive tt-axis and converges under our assumptions on f(z)f(z).

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Remark.
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H. Weyl.

Problem 1.151

The Maclaurin series of e−ze^{-z}, log⁡(1+z)\log (1+z) and (1+z)−p(1+z)^{-p}, p>0p>0, envelop the respective functions for ℜz≥0\Re z \geq 0, z≠0z \neq 0.

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Remark.
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For e−ze^{-z} cf.E. Landau: Arch. Math. Phys. Ser. 3, Vol. 24, p. 104 (1915). Cf.problem 150.

Problem 1.152

Let zz be restricted by the following conditions:

−π4≤arg⁡z≤π4,3π4≤arg⁡z≤5π4,z≠0. -\frac{\pi }{4} \leq \arg z \leq \frac{\pi }{4}, \qquad \frac{3\pi }{4} \leq \arg z \leq \frac{5\pi }{4}, \qquad z \neq 0 .

Then the function ez2/2∫z∞e−t2/2 dte^{z^2/2} \displaystyle \int_z^{\infty } e^{-t^2/2} \, dt is enveloped by the series

1z−1z3+1⋅3z5−1⋅3⋅5z7+⋯ \frac{1}{z} - \frac{1}{z^3} + \frac{1 \cdot 3}{z^5} - \frac{1 \cdot 3 \cdot 5}{z^7} + \cdots

(strictly enveloped if zz is real).

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Problem 1.153

Suppose that ana_n and bnb_n are arbitrary complex numbers that have the same argument, i.e.anbn\dfrac {a_n}{b_n} is real and positive. If at a certain point z≠0z \neq 0 the two series

a0+a1z+a2z2+⋯+anzn+⋯ ,b0+b1z+b2z2+⋯+bnzn+⋯ a_0+a_1z+a_2z^2+\cdots +a_nz^n+\cdots , \qquad b_0+b_1z+b_2z^2+\cdots +b_nz^n+\cdots

envelop the values φ(z)\varphi (z) and ψ(z)\psi (z) then the combined series

a0+b0+(a1+b1)z+(a2+b2)z2+⋯+(an+bn)zn+⋯ a_0+b_0+(a_1+b_1)z+(a_2+b_2)z^2+\cdots +(a_n+b_n)z^n+\cdots

envelops the value φ(z)+ψ(z)\varphi (z)+\psi (z) for this zz. (The same is true for enveloping in the strict sense if all the coefficients and zz are real.)

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Problem 1.154

If zz lies in the sectors described in problem 152 the function zcoth⁡zz \coth z is enveloped by its power series

zcoth⁡z=zez+e−zez−e−z=1+B1(2z)22!−B2(2z)44!+B3(2z)66!+⋯ z \coth z = z \frac{e^z+e^{-z}}{e^z-e^{-z}} = 1 + B_1 \frac{(2z)^2}{2!} - B_2 \frac{(2z)^4}{4!} + B_3 \frac{(2z)^6}{6!} + \cdots

(strictly enveloped if zz is real). The coefficients B1,B2,B3,…B_1, B_2, B_3, \ldots are called the Bernoulli numbers.

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Remark.
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Cauchy: C. R. Acad. Sci. (Paris) Sér. A-B, Vol. 17, pp. 370-376 (1843); cf.also E. T. Whittaker and G. N. Watson, p. 136, ex. 7.

Problem 1.155

The function

ω(z)=log⁡Γ(1+z)−(z+12)log⁡z+z−12log⁡(2π)[see Problem 31 of Part II] \omega (z) = \log \Gamma (1+z) - \left(z+\frac{1}{2}\right) \log z + z - \frac{1}{2} \log (2\pi ) \qquad \text{[see Problem 31 of Part II]}

can, for ℜz>0\Re z>0, be written as an integral

ω(z)=2∫0∞arctan⁡tze2πt−1 dt. \omega (z) = 2 \int _0^{\infty } \frac{\arctan \dfrac {t}{z}}{e^{2\pi t}-1} \, dt .

[Cf.E.T.Whittaker and G.N.Watson, pp.251-252.] Prove that the resulting (divergent) Stirling series

B11⋅2⋅z−B23⋅4⋅z3+B35⋅6⋅z5−⋯ \frac{B_1}{1 \cdot 2 \cdot z} - \frac{B_2}{3 \cdot 4 \cdot z^3} + \frac{B_3}{5 \cdot 6 \cdot z^5} - \cdots

envelops the function ω(z)\omega (z) if ℜz>0\Re z>0 and −π4≤arg⁡z≤π4-\dfrac {\pi }{4} \leq \arg z \leq \dfrac {\pi }{4}.

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Remark.
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Cf.Cauchy, l.c.solution 154.

Problem 1.156

Assume that φ(x)\varphi (x) is defined for positive xx and that for xx large enough it can be represented in the form

φ(x)=a0+a1x+a2x2+⋯+akxk+⋯ \varphi (x) = a_0 + \frac{a_1}{x} + \frac{a_2}{x^2} + \cdots + \frac{a_k}{x^k} + \cdots

aka_k real, k=0,1,2,…k=0,1,2,\ldots The infinite series

φ(1)+φ(2)+φ(3)+⋯+φ(n)+⋯ \varphi (1)+\varphi (2)+\varphi (3)+\cdots +\varphi (n)+\cdots

converges if and only if a0=a1=0a_0=a_1=0.

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Problem 1.157

(continued). Suppose φ(n)≠0\varphi (n) \neq 0. The infinite product

φ(1)φ(2)φ(3)⋯φ(n)⋯ \varphi (1)\varphi (2)\varphi (3) \cdots \varphi (n) \cdots

converges if and only if a0=1a_0=1, a1=0a_1=0.

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Problem 1.158

(continued). Under what conditions does the following infinite series converge

φ(1)+φ(1)φ(2)+φ(1)φ(2)φ(3)+⋯+φ(1)φ(2)φ(3)⋯φ(n)+⋯ ? \varphi (1) + \varphi (1)\varphi (2) + \varphi (1)\varphi (2)\varphi (3) + \cdots + \varphi (1)\varphi (2)\varphi (3) \cdots \varphi (n) + \cdots \, ?
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Problem 1.159

For which positive values of α\alpha does the following series converge

∑n=1∞(2−eα)(2−eα/2)⋯(2−eα/n) ? \sum _{n=1}^{\infty } (2-e^{\alpha })(2-e^{\alpha /2}) \cdots (2-e^{\alpha /n}) \, ?
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Problem 1.160
∫01x−x dx=∑n=1∞n−n. \int _0^1 x^{-x} \, dx = \sum _{n=1}^{\infty } n^{-n} .
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Problem 1.161

Considering positive square roots only we find

1+1+1+⋯=1+11+11+⋱ . \sqrt{1+\sqrt{1+\sqrt{1+\cdots }}} = 1 + \cfrac {1}{1+\cfrac {1}{1+\ddots }} \, .
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Problem 1.162

Let a1,a2,…,an,…a_1, a_2, \ldots , a_n, \ldots be positive numbers and put

tn=a1+a2+⋯+an , t_n = \sqrt{a_1+\sqrt{a_2+\cdots +\sqrt{a_n}}} \, ,

where only positive square roots are considered. The convergence of the sequence

(t)t1,t2,t3,…,tn,… (t) \qquad t_1, t_2, t_3, \ldots , t_n, \ldots

is related to

lim sup⁡n→∞log⁡log⁡ann=α \limsup _{n \rightarrow \infty } \frac{\log \log a_n}{n} = \alpha

in the following way:

if α<2 then (t) converges, \text{if } \alpha <2 \text{ then } (t) \text{ converges,} if α>2 then (t) diverges. \text{if } \alpha >2 \text{ then } (t) \text{ diverges.}
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Remark.
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G. Pólya, Problem: Arch. Math. Phys. Ser. 3, Vol. 24, p. 84 (1916). Solved by G. Szegö: Arch. Math. Phys. Ser. 3, Vol. 25, pp. 88-89 (1917).

Problem 1.163

(continued). The sequence (t)(t) is certainly convergent if the series

∑n=1∞2−nan(a1a2⋯an)−12 converges. \sum _{n=1}^{\infty } 2^{-n} a_n (a_1 a_2 \cdots a_n)^{-\frac{1}{2}} \text{ converges.}
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Problem 1.164

We have for 0<q<10<q<1

1−q1+q(1−q21+q2)12(1−q41+q4)14(1−q81+q8)18⋯=(1−q)2. \frac{1-q}{1+q} \left(\frac{1-q^2}{1+q^2}\right)^{\frac{1}{2}} \left(\frac{1-q^4}{1+q^4}\right)^{\frac{1}{4}} \left(\frac{1-q^8}{1+q^8}\right)^{\frac{1}{8}} \cdots = (1-q)^2 .
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Remark.
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Jacobi, l.c.solution 53, §52, Corollarium: Werke, Vol. 1, pp. 200-201.

Problem 1.165

The series

⋯+f′′(x)+f′(x)+f(x)+∫0xf(ξ1) dξ1+∫0xdξ1∫0ξ1f(ξ2) dξ2+⋯ \cdots + f''(x) + f'(x) + f(x) + \int _0^x f(\xi _1) \, d\xi _1 + \int _0^x d\xi _1 \int _0^{\xi _1} f(\xi _2) \, d\xi _2 + \cdots

is infinite in two directions. Suppose it does converge uniformly on some interval. Which function does the series represent?

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Problem 1.166

Let φn(x)\varphi_n(x) and ψn(x)\psi_n(x) denote the polynomials of degree nn, n=0,1,2,…n=0,1,2,\ldots defined by the formulas

φ0(x)=1,φn′(x)=φn−1(x),φn(0)=0, \varphi _0(x)=1, \quad \varphi _n'(x)=\varphi _{n-1}(x), \quad \varphi _n(0)=0 , ψ0(x)=1,ψn(x+1)−ψn(x)=ψn−1(x),ψn(0)=0,n=1,2,3,… \psi _0(x)=1, \quad \psi _n(x+1)-\psi _n(x)=\psi _{n-1}(x), \quad \psi _n(0)=0, \qquad n=1,2,3,\ldots

Find the two sums φ(x)\varphi (x) and ψ(x)\psi (x),

φ(x)=φ0(x)+φ1(x)+⋯+φn(x)+⋯ , \varphi (x) = \varphi _0(x)+\varphi _1(x)+\cdots +\varphi _n(x)+\cdots , ψ(x)=ψ0(x)+ψ1(x)+⋯+ψn(x)+⋯ . \psi (x) = \psi _0(x)+\psi _1(x)+\cdots +\psi _n(x)+\cdots .
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Problem 1.167

We define

xn=yne−112n,yn=n! n−n−12en,n=1,2,3,… x_n = y_n e^{-\frac{1}{12n}}, \qquad y_n = n! \, n^{-n-\frac{1}{2}} e^n, \qquad n=1,2,3,\ldots

Then each interval (xk,yk)(x_k,y_k), k=1,2,3,…k=1,2,3,\ldots, contains the interval (xk+1,yk+1)(x_{k+1},y_{k+1}) as a subinterval.

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Problem 1.168

The sequence

an=(1+1n)n+p,n=1,2,3,… a_n = \left(1+\frac{1}{n}\right)^{n+p}, \qquad n=1,2,3,\ldots

is monotone decreasing if and only if p≥12p \geq \dfrac {1}{2}.

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Remark.
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I. Schur.

Problem 1.169

The sequence

an=(1+1n)n(1+xn),n=1,2,3,… a_n = \left(1+\frac{1}{n}\right)^n \left(1+\frac{x}{n}\right), \qquad n=1,2,3,\ldots

is monotone decreasing if and only if x≥12x \geq \dfrac {1}{2}.

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Problem 1.170

Let nn be a positive integer. Then we have

e2n+2<e−(1+1n)n<e2n+1. \frac{e}{2n+2} < e - \left(1+\frac{1}{n}\right)^n < \frac{e}{2n+1} .
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Remark.
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Cf.problem No. 1098, Nouv. Annls Math. Ser. 2, Vol. 11, p. 480 (1872). Solved by C. Moreau; Nouv. Annls Math. Ser. 2, Vol. 13, p. 61 (1874).

Problem 1.171

As is well-known, the number e=lim⁡n→∞(1+1n)ne = \lim_{n \rightarrow \infty } \left(1+\dfrac {1}{n}\right)^n is contained in the interval

(1+1n)n<e<(1+1n)n+1[problem 168] \left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1} \qquad \text{[problem 168]}

In which quarter of the interval is it contained?

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Remark.
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I. Schur.

Problem 1.172

The sequence

an=(1+xn)n+1,n=1,2,3,… a_n = \left(1+\frac{x}{n}\right)^{n+1}, \qquad n=1,2,3,\ldots

is monotone decreasing if and only if 0<x≤20<x \leq 2.

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Remark.
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I. Schur.

Problem 1.173

Prove that the nn-times iterated sine function

sin⁡nx=sin⁡(sin⁡n−1x),sin⁡1x=sin⁡x, \sin _n x = \sin (\sin _{n-1} x), \qquad \sin _1 x = \sin x ,

converges to 00 as n→∞n \rightarrow \infty if sin⁡x>0\sin x>0 and that, moreover,

lim⁡n→∞n3sin⁡nx=1. \lim _{n \rightarrow \infty } \sqrt{\frac{n}{3}} \sin _n x = 1 .
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Remark.
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Proof based on a communication of E. Jacobsthal.

Problem 1.174

Assume that 0<f(x)<x0<f(x)<x and

f(x)=x−axk+bxl+xlε(x),lim⁡x→0ε(x)=0 f(x) = x - ax^k + bx^l + x^l \varepsilon (x), \qquad \lim _{x \rightarrow 0} \varepsilon (x) = 0

for 0<x<x00<x<x_0, where 1<k<l1<k<l and a,ba,b positive. Put

v0=x,v1=f(v0),v2=f(v1),…,vn=f(vn−1),… v_0=x, \quad v_1=f(v_0), \quad v_2=f(v_1), \quad \ldots , \quad v_n=f(v_{n-1}), \quad \ldots

Then we have for n→∞n \rightarrow \infty

n1k−1vn→[(k−1)a]−1k−1. n^{\frac{1}{k-1}} v_n \rightarrow \big[(k-1)a\big]^{-\frac{1}{k-1}} .
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Problem 1.175

Discuss the convergence of the series

v1s+v2s+v3s+⋯ , v_1^s + v_2^s + v_3^s + \cdots ,

where

v1=sin⁡x,v2=sin⁡sin⁡x,…,vn=sin⁡vn−1,… v_1 = \sin x, \quad v_2 = \sin \sin x, \quad \ldots , \quad v_n = \sin v_{n-1}, \quad \ldots

We can obviously assume that v1>0v_1>0.

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Remark.
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J. Ouspensky, Problem: Arch. Math. Phys. Ser. 3, Vol. 20, p. 83 (1913).

Problem 1.176

Prove the formula

ex−1=u1+u1u2+u1u2u3+⋯ e^x - 1 = u_1 + u_1 u_2 + u_1 u_2 u_3 + \cdots

where

u1=x≥0,un+1=log⁡eun−1un,n=1,2,3,… u_1 = x \geq 0, \qquad u_{n+1} = \log \frac{e^{u_n}-1}{u_n}, \qquad n=1,2,3,\ldots
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Remark.
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Cf.E. Cesàro, Problem: Nouv. Annls Math. Ser. 3, Vol. 7, p. 400 (1888). Solved by Audibert: Nouv. Annls Math. Ser. 3, Vol. 11, p. 35 (1892).

Problem 1.177

Compute

s=cos⁡3φ−13cos⁡33φ+132cos⁡332φ−133cos⁡333φ+⋯ . s = \cos ^3 \varphi - \frac{1}{3} \cos ^3 3\varphi + \frac{1}{3^2} \cos ^3 3^2\varphi - \frac{1}{3^3} \cos ^3 3^3 \varphi + \cdots .
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Remark.
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C. A. Laisant, Problem: Nouv. Annls Math. Ser. 2, Vol. 9, p. 144 (1870). Solved by H. Rumpen: Nouv. Annls Math. Ser. 2, Vol. 11, p. 232 (1872).

Problem 1.178

Let ana_n, bnb_n, bn≠0b_n \neq 0, n=0,1,2,3,…n=0,1,2,3,\ldots, be two sequences that satisfy the conditions:

  1. the power series f(x)=∑n=0∞anxnf(x) = \sum_{n=0}^{\infty } a_n x^n has a positive radius of convergence rr.

  2. the limit

    lim⁡n→∞bnbn+1=q \lim _{n \rightarrow \infty } \frac{b_n}{b_{n+1}} = q

    exists and ∣q∣<r\left|q\right|<r.

We define

cn=a0bn+a1bn−1+⋯+anb0,n=0,1,2,… c_n = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0, \qquad n=0,1,2,\ldots

Then cnbn\dfrac {c_n}{b_n} converges to f(q)f(q) as n→∞n \rightarrow \infty.

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Remark.
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I. Schur, Problem: Arch. Math. Phys. Ser. 3, Vol. 27, p. 162 (1918). Cf.O. Szász: Sber. Berlin Math. Ges. Vol. 21, pp. 25-29 (1922).

Problem 1.179

Let

fn(x)=an1x+an2x2+an3x3+⋯ ,n=1,2,3,… f_n(x) = a_{n1}x + a_{n2}x^2 + a_{n3}x^3 + \cdots , \qquad n=1,2,3,\ldots

be arbitrary functions, ∣ank∣<A\left|a_{nk}\right|<A for all positive integral values of nn and kk and

lim⁡n→∞fn(x)=0if0<x<1. \lim _{n \rightarrow \infty } f_n(x) = 0 \quad \text{if} \quad 0<x<1 .

Then we have for fixed kk, k=1,2,3,…k=1,2,3,\ldots

lim⁡n→∞ank=0. \lim _{n \rightarrow \infty } a_{nk} = 0 .
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Remark.
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Special case of an important proposition in function theory by Vitali. Cf.E. Lindelöf: Bull. Soc. Math. France Vol. 41, p. 171 (1913).

Problem 1.180

Suppose that the series

an0+an1+an2+⋯+ank+⋯=sn,n=0,1,2,…, a_{n0}+a_{n1}+a_{n2}+\cdots +a_{nk}+\cdots = s_n, \qquad n=0,1,2,\ldots ,

have a common convergent majorant

A0+A1+A2+⋯+Ak+⋯=S, A_0+A_1+A_2+\cdots +A_k+\cdots = S ,

i.e.for each kk the inequalities ∣ank∣≤Ak\left|a_{nk}\right| \leq A_k hold simultaneously for all nn. Assume furthermore that

lim⁡n→∞ank=ak \lim _{n \rightarrow \infty } a_{nk} = a_k

exists for k=0,1,2,…k=0,1,2,\ldots Then the series

a0+a1+a2+⋯+ak+⋯=s a_0+a_1+a_2+\cdots +a_k+\cdots = s

is convergent and

lim⁡n→∞sn=s. \lim _{n \rightarrow \infty } s_n = s .
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Problem 1.181

Justify the limits in problems 53 and 59.

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Problem 1.181.1

Assume that

ai1+ai2+⋯+ain+⋯=si a_{i1}+a_{i2}+\cdots +a_{in}+\cdots = s_i

converges for i=1,2,3,…i=1,2,3,\ldots, define UiU_i as the least upper bound of

∣ai1+ai2+⋯+ain∣,n=1,2,3,…, \left|a_{i1}+a_{i2}+\cdots +a_{in}\right|, \qquad n=1,2,3,\ldots ,

and assume that

U1+U2+⋯+Un+⋯ U_1+U_2+\cdots +U_n+\cdots

converges. Then the series

(∗)a11+a12+a21+a13+a22+a31+⋯+a1n+a2,n−1+⋯ , (*) \qquad a_{11}+a_{12}+a_{21}+a_{13}+a_{22}+a_{31}+\cdots +a_{1n}+a_{2,n-1}+\cdots ,

which you obtain by arranging the numbers in the array

a11a12a13⋯a1n⋯a21a22a23⋯a2,n−1⋯a31a32a33⋯⋯⋯⋯⋯⋯ \begin{matrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} & \cdots \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2,n-1} & \cdots \\ a_{31} & a_{32} & a_{33} & \cdots & & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \end{matrix}

``diagonally'', converges and its sum is

s1+s2+⋯+sn+⋯ . s_1+s_2+\cdots +s_n+\cdots .
?
Remark.
?

The interesting point is that the absolute convergence of the double series ∑∑aik\sum \sum a_{ik} is not assumed.

Problem 1.181.1

Assume that

ai1+ai2+⋯+ain+⋯=si a_{i1}+a_{i2}+\cdots +a_{in}+\cdots = s_i

converges for i=1,2,3,…i=1,2,3,\ldots, define UiU_i as the least upper bound of

∣ai1+ai2+⋯+ain∣,n=1,2,3,…, \left|a_{i1}+a_{i2}+\cdots +a_{in}\right|, \qquad n=1,2,3,\ldots ,

and assume that

U1+U2+⋯+Un+⋯ U_1+U_2+\cdots +U_n+\cdots

converges. Then the series

(∗)a11+a12+a21+a13+a22+a31+⋯+a1n+a2,n−1+⋯ , (*) \qquad a_{11}+a_{12}+a_{21}+a_{13}+a_{22}+a_{31}+\cdots +a_{1n}+a_{2,n-1}+\cdots ,

which you obtain by arranging the numbers in the array

a11a12a13⋯a1n⋯a21a22a23⋯a2,n−1⋯a31a32a33⋯⋯⋯⋯⋯⋯ \begin{matrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} & \cdots \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2,n-1} & \cdots \\ a_{31} & a_{32} & a_{33} & \cdots & & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \end{matrix}

``diagonally'', converges and its sum is

s1+s2+⋯+sn+⋯ . s_1+s_2+\cdots +s_n+\cdots .
?
Remark.
?

The interesting point is that the absolute convergence of the double series ∑∑aik\sum \sum a_{ik} is not assumed.

Problem 1.182

If α\alpha is fixed, α>0\alpha >0, and nn an integer increasing to +∞+\infty, then

∑ν=1∞′ να−1(nα−να)−2∼n1−α3(πα)2. {\sum _{\nu =1}^{\infty }}' \, \nu ^{\alpha -1} (n^{\alpha }-\nu ^{\alpha })^{-2} \sim \frac{n^{1-\alpha }}{3} \left(\frac{\pi }{\alpha }\right)^2 .

The meaningless term with subscript ν=n\nu =n has to be omitted, which is indicated by the comma at the summation sign. Notice the case α=1\alpha =1.

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Problem 1.183

Let the numbers ε0,ε1,ε2,…,εn,…\varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots , \varepsilon_n, \ldots assume one of the three values −1,0,1-1, 0, 1. Then we have

ε02+ε12+ε22+⋯=2sin⁡(π4∑n=0∞ε0ε1ε2⋯εn2n). \varepsilon _0 \sqrt{2+\varepsilon _1 \sqrt{2+\varepsilon _2 \sqrt{2+\cdots }}} = 2 \sin \left( \frac{\pi }{4} \sum _{n=0}^{\infty } \frac{\varepsilon _0 \varepsilon _1 \varepsilon _2 \cdots \varepsilon _n}{2^n} \right) .
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Remark.
?

The left hand side must be interpreted as the limit of

ε02+ε12+ε22+⋯+εn2 ,n=0,1,2,… \varepsilon _0 \sqrt{2+\varepsilon _1 \sqrt{2+\varepsilon _2 \sqrt{2+\cdots +\varepsilon _n \sqrt{2}}}} \, , \qquad n=0,1,2,\ldots

for n→∞n \rightarrow \infty. These expressions are well defined for all nn. Non-negative square roots are used throughout.

Problem 1.184

Every value xx of the interval [−2,2][-2,2] can be written in the form

x=ε02+ε12+ε22+⋯ x = \varepsilon _0 \sqrt{2+\varepsilon _1 \sqrt{2+\varepsilon _2 \sqrt{2+\cdots }}}

where εk\varepsilon_k, k=0,1,2,…k=0,1,2,\ldots, is either −1-1 or +1+1. The representation is unique if xx is not of the form 2cos⁡p2qπ2 \cos \dfrac {p}{2^q} \pi, p,qp, q integers, 0<p<2q0<p<2^q. These are the only numbers that may be written in the finite form

ε02+ε12+ε22+⋯+εn2 . \varepsilon _0 \sqrt{2+\varepsilon _1 \sqrt{2+\varepsilon _2 \sqrt{2+\cdots +\varepsilon _n \sqrt{2}}}} \, .

The finite representation can be extended to an infinite one in two ways: by putting either

εn+1=1,εn+2=−1,εn+3=εn+4=⋯=1, \varepsilon _{n+1}=1, \quad \varepsilon _{n+2}=-1, \quad \varepsilon _{n+3}=\varepsilon _{n+4}=\cdots =1 ,

or

εn+1=−1,εn+2=−1,εn+3=εn+4=⋯=1. \varepsilon _{n+1}=-1, \quad \varepsilon _{n+2}=-1, \quad \varepsilon _{n+3}=\varepsilon _{n+4}=\cdots =1 .
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Remark.
?

Cf.S. Pincherle: Atti Accad. Sci. Torino, Cl. Sci. Fis. Mat. Natur. Vol. 53, pp. 745-763 (1917-1918); Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. Ser. 5, Vol. 27, 2nd Sem. pp. 177-183 (1918).

Problem 1.185

(continued). The number xx is of the form x=2cos⁡kπx=2\cos k\pi, kk rational, if and only if the sequence ε0,ε1,ε2,…\varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots is periodical after a certain term.

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Problem 1.185.1

Construct a sequence of real numbers a1,a2,…,an,…a_1, a_2, \ldots , a_n, \ldots so that the series

a1l+a2l+⋯+anl+⋯ a_1^l + a_2^l + \cdots + a_n^l + \cdots

diverges for l=5l=5, but converges when ll is any odd positive integer different from 5.

?
Remark.
?

For precursory heuristic considerations see MD, Vol. 2, pp. 49-50 and 171.

Problem 1.185.1

Construct a sequence of real numbers a1,a2,…,an,…a_1, a_2, \ldots , a_n, \ldots so that the series

a1l+a2l+⋯+anl+⋯ a_1^l + a_2^l + \cdots + a_n^l + \cdots

diverges for l=5l=5, but converges when ll is any odd positive integer different from 5.

?
Remark.
?

For precursory heuristic considerations see MD, Vol. 2, pp. 49-50 and 171.

Problem 1.185.2

(continued). Let the set of all odd positive integers be divided (arbitrarily) into two complementary subsets CC and DD (having no element in common). Show that there exists a sequence of real numbers a1,a2,…,an,…a_1, a_2, \ldots , a_n, \ldots such that the series mentioned converges when ll belongs to CC and diverges when ll belongs to DD.

?
Remark.
?

G. Pólya, Problem: Amer. Math. Monthly Vol. 51, p. 593 (1944). Solved by N. J. Fine: Amer. Math. Monthly Vol. 53, pp. 283-284 (1946).

Problem 1.185.2

(continued). Let the set of all odd positive integers be divided (arbitrarily) into two complementary subsets CC and DD (having no element in common). Show that there exists a sequence of real numbers a1,a2,…,an,…a_1, a_2, \ldots , a_n, \ldots such that the series mentioned converges when ll belongs to CC and diverges when ll belongs to DD.

?
Remark.
?

G. Pólya, Problem: Amer. Math. Monthly Vol. 51, p. 593 (1944). Solved by N. J. Fine: Amer. Math. Monthly Vol. 53, pp. 283-284 (1946).

Problem 1.186

Tabulate the numbers SknS_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

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Remark.
?

List concretely the different possibilities for small nn, use problem 187 for all n≥3n \geq 3.

Problem 1.187

Obviously

S1n=Snn=1. S_1^n = S_n^n = 1 .

Show that

Skn+1=Sk−1n+kSkn. S_k^{n+1} = S_{k-1}^n + k S_k^n .
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Problem 1.188

Show that

Skkzk+1+Skk+1zk+2+⋯+Sknzn+1+⋯=1z(z−1)(z−2)⋯(z−k). \frac{S_k^k}{z^{k+1}} + \frac{S_k^{k+1}}{z^{k+2}} + \cdots + \frac{S_k^n}{z^{n+1}} + \cdots = \frac{1}{z(z-1)(z-2) \cdots (z-k)} .
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Problem 1.189

Show that, for n≥1n \geq 1,

Skn=1k![kn−(k1)(k−1)n+(k2)(k−2)n−⋯+(−1)k0n]. S_k^n = \frac{1}{k!} \left[ k^n - \binom {k}{1}(k-1)^n + \binom {k}{2}(k-2)^n - \cdots + (-1)^k 0^n \right] .
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Problem 1.190

Show that

∑n=k∞Sknznn!=(ez−1)kk!. \sum _{n=k}^{\infty } \frac{S_k^n z^n}{n!} = \frac{(e^z-1)^k}{k!} .
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Problem 1.191

Prove the identity in xx

xn=S1nx+S2nx(x−1)+⋯+Snnx(x−1)⋯(x−n+1).[Problem 189; see also Problem 221 of Part III.] x^n = S_1^n x + S_2^n x(x-1) + \cdots + S_n^n x(x-1) \cdots (x-n+1) . \qquad \text{[Problem 189; see also Problem 221 of Part III.]}
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Problem 1.192

Prove the identity of problem 191 independently of problem 189 and derive hence a new proof for problem 189.

?
Remark.
?

By a combinatorial argument.

Problem 1.193

Define T0=1T_0=1. Then

T0+T1z1!+T2z22!+⋯+Tnznn!+⋯=eez−1. T_0 + \frac{T_1 z}{1!} + \frac{T_2 z^2}{2!} + \cdots + \frac{T_n z^n}{n!} + \cdots = e^{e^z-1} .
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Problem 1.194

Show that

Tn+1=(n0)Tn+(n1)Tn−1+(n2)Tn−2+⋯+(nn)T0. T_{n+1} = \binom {n}{0} T_n + \binom {n}{1} T_{n-1} + \binom {n}{2} T_{n-2} + \cdots + \binom {n}{n} T_0 .
?
Remark.
?

By a combinatorial argument.

Problem 1.195

Use problem 194 to prove problem 193 [see also problem 34].

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Problem 1.196

Show that for n≥1n \geq 1

Tn=1e(1n1!+2n2!+3n3!+⋯ ). T_n = \frac{1}{e} \left( \frac{1^n}{1!} + \frac{2^n}{2!} + \frac{3^n}{3!} + \cdots \right) .
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Problem 1.197

Tabulate the numbers skns_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

?
Remark.
?

List concretely the different possibilities for small nn, use problem 198 for all n≥3n \geq 3.

Problem 1.198

Obviously

s1n=(n−1)!,snn=1. s_1^n = (n-1)!, \qquad s_n^n = 1 .

Show that

skn+1=sk−1n+nskn. s_k^{n+1} = s_{k-1}^n + n s_k^n .
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Problem 1.199

Show that

x(x+1)(x+2)⋯(x+n−1)=s1nx+s2nx2+⋯+snnxn x(x+1)(x+2) \cdots (x+n-1) = s_1^n x + s_2^n x^2 + \cdots + s_n^n x^n

or, which is the same, that

x(x−1)(x−2)⋯(x−n+1)=(−1)n−1s1nx+⋯−sn−1nxn−1+snnxn. x(x-1)(x-2) \cdots (x-n+1) = (-1)^{n-1} s_1^n x + \cdots - s_{n-1}^n x^{n-1} + s_n^n x^n .
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Problem 1.200

Show that

∑n=k∞sknznn!=1k!(log⁡11−z)k. \sum _{n=k}^{\infty } \frac{s_k^n z^n}{n!} = \frac{1}{k!} \left( \log \frac{1}{1-z} \right)^k .
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Remark.
?

Compare problem 200 with problem 190, problem 199 with problem 188, and again problem 199 with problem 191. See also Problem 54.2 of Part VII and Problem 247.1 of Part VIII.

Problem 1.201

Tabulate the numbers S~kn\widetilde{S}_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

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Remark.
?

List concretely the different possibilities for small nn, use problem 202 for all n≥4n \geq 4.

Problem 1.202

Obviously

S~1n=1 when n≥2,S~kn=0 when n<2k. \widetilde{S}_1^n = 1 \text{ when } n \geq 2, \qquad \widetilde{S}_k^n = 0 \text{ when } n<2k .

Show that

S~kn+1=nS~k−1n−1+kS~kn. \widetilde{S}_k^{n+1} = n \widetilde{S}_{k-1}^{n-1} + k \widetilde{S}_k^n .
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Problem 1.203

Show that

S~2n=2n−1−n−1,S~n2n=1⋅3⋅5⋯(2n−1). \widetilde{S}_2^n = 2^{n-1}-n-1, \qquad \widetilde{S}_n^{2n} = 1 \cdot 3 \cdot 5 \cdots (2n-1) .
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Problem 1.204

Returning to SknS_k^n, show that

Sn−1n=(n2),Sn−2n=(n3)3n−54,Sn−3n=(n4)(n−2)(n−3)2 S_{n-1}^n = \binom {n}{2}, \qquad S_{n-2}^n = \binom {n}{3} \frac{3n-5}{4}, \qquad S_{n-3}^n = \binom {n}{4} \frac{(n-2)(n-3)}{2}

and, generally, that Sn−anS_{n-a}^n is a polynomial in nn of degree 2a2a divisible by n(n−1)(n−2)⋯(n−a)n(n-1)(n-2) \cdots (n-a); of course, a≥1a \geq 1.

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Problem 1.205

Define T~0=1\widetilde{T}_0=1,

T~n=S~1n+S~2n+⋯+S~nn \widetilde{T}_n = \widetilde{S}_1^n + \widetilde{S}_2^n + \cdots + \widetilde{S}_n^n

for n≥1n \geq 1. Compute TnT_n and T~n\widetilde{T}_n for n≤8n \leq 8.

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Problem 1.206
T~n+1=(n1)T~n−1+(n2)T~n−2+⋯+(nn−1)T~1+(nn)T~0. \widetilde{T}_{n+1} = \binom {n}{1} \widetilde{T}_{n-1} + \binom {n}{2} \widetilde{T}_{n-2} + \cdots + \binom {n}{n-1} \widetilde{T}_1 + \binom {n}{n} \widetilde{T}_0 .
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Problem 1.207
T~0+T~1z1!+T~2z22!+⋯+T~nznn!+⋯=eez−1−z. \widetilde{T}_0 + \frac{\widetilde{T}_1 z}{1!} + \frac{\widetilde{T}_2 z^2}{2!} + \cdots + \frac{\widetilde{T}_n z^n}{n!} + \cdots = e^{e^z-1-z} .
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Problem 1.208
T~n=(n0)Tn−(n1)Tn−1+(n2)Tn−2−⋯+(−1)n(nn)T0. \widetilde{T}_n = \binom {n}{0} T_n - \binom {n}{1} T_{n-1} + \binom {n}{2} T_{n-2} - \cdots + (-1)^n \binom {n}{n} T_0 .
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Remark.
?

T~n=ΔnT0\widetilde{T}_n = \Delta^n T_0, if we use the notation of the calculus of finite differences, see the introduction to Problem 220 of Part III.

Problem 1.209

If the function F(t)F(t) has an nn-th derivative

(ddx)nF(ex)=S1nF′(ex)ex+S2nF′′(ex)e2x+⋯+SnnF(n)(ex)enx. \left( \frac{d}{dx} \right)^n F(e^x) = S_1^n F'(e^x) e^x + S_2^n F''(e^x) e^{2x} + \cdots + S_n^n F^{(n)}(e^x) e^{nx} .
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Problem 1.210

Derive the identities in the variables zz and ww:

∑n=0∞∑k≤n(nk)wkznn!=ez(w+1),(1) \sum _{n=0}^{\infty } \sum _{k \leq n} \binom {n}{k} \frac{w^k z^n}{n!} = e^{z(w+1)} , \tag {1} ∑n=0∞∑k≤nSknwkznn!=e(ez−1)w,(2) \sum _{n=0}^{\infty } \sum _{k \leq n} S_k^n \frac{w^k z^n}{n!} = e^{(e^z-1)w} , \tag {2} ∑n=0∞∑k≤nsknwkznn!=(1−z)−w(3) \sum _{n=0}^{\infty } \sum _{k \leq n} s_k^n \frac{w^k z^n}{n!} = (1-z)^{-w} \tag {3}

either on the basis of the foregoing [problems 10, 190, 200] or independently of the foregoing.

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Remark.
?

On the other hand, we could regard (1), (2) and (3) as defining (nk)\binom {n}{k}, SknS_k^n and skns_k^n, respectively, and then take them as basis and starting point for establishing the propositions, especially the combinatorial propositions, discussed in the section herewith concluded. For additional material on the subject treated in this section see Problem 62.1 of Part V, Problem 54.2 of Part VII, Problems 22.1, 22.2, 22.3, 58.3, 247.1 of Part VIII, and Problems 191.1, 191.2 of this Part.