Abelian theorems

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Prove the following statement: lim⁡x→1(x1+x−x21+x2+x31+x3−…)=14\lim_{x \rightarrow 1}\left(\frac{x}{1+x}-\frac{x^{2}}{1+x^{2}}+\frac{x^{3}}{1+x^{3}}-\ldots \right) = \frac{1}{4}.

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Prove the following statement: if an>0a_{n} > 0, bn>0b_{n} > 0, the series ∑anxn\sum a_{n}x^{n} is divergent at x=1x = 1, and convergent for ∣x∣<1\left|x\right| < 1, and anbn→A\frac{a_{n}}{b_{n}} \rightarrow A as n→∞n \rightarrow \infty, then

lim⁡x→1−0∑anxn∑bnxn=A. \lim _{x \rightarrow 1-0} \frac{\sum a_{n}x^{n}}{\sum b_{n}x^{n}} = A.
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Prove the following statement: if ∑an\sum a_{n} diverges and AnBn→A\frac{A_{n}}{B_{n}} \rightarrow A, where An=∑ν=1naνA_{n} = \sum_{\nu =1}^{n} a_{\nu }, Bn=∑ν=1nbνB_{n} = \sum_{\nu =1}^{n} b_{\nu }, then

lim⁡x→1−0∑anxn∑bnxn=A. \lim _{x \rightarrow 1-0} \frac{\sum a_{n}x^{n}}{\sum b_{n}x^{n}} = A.
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Prove the following statement: if annp−1→a\frac{a_{n}}{n^{p-1}} \rightarrow a as n→∞n \rightarrow \infty, then

lim⁡x→1−0(1−x)p∑n=1∞anxn=aΓ(p). \lim _{x \rightarrow 1-0} (1-x)^{p}\sum _{n=1}^{\infty } a_{n}x^{n} = a\Gamma (p).
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Prove the following statement: if lim⁡n→∞a1+a2+…+annp=a\lim_{n \rightarrow \infty } \frac{a_{1}+a_{2}+\ldots +a_{n}}{n^{p}} = a, then

lim⁡x→1−0(1−x)p∑n=1∞anxn=aΓ(p+1);p>−1. \lim _{x \rightarrow 1-0} (1-x)^{p}\sum _{n=1}^{\infty } a_{n}x^{n} = a\Gamma (p+1); \quad p > -1.
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Prove the following statement (Borel's theorem): if the series ∑an\sum a_{n} converges and has sum ss, and moreover

∑n=0∞anxnn!=f(x),then∫0∞e−xf(x) dx=s. \sum _{n=0}^{\infty } \frac{a_{n}x^{n}}{n!} = f(x), \quad \text{then} \quad \int _{0}^{\infty } e^{-x}f(x)\, dx = s.
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A necessary and sufficient condition that

∑n=0∞∣αn∣2<∞ \sum _{n=0}^{\infty } \left|\alpha _{n}\right|^{2} < \infty

is that the numbers

∑n=0∞∣αn∣2r2n(0<r<1) \sum _{n=0}^{\infty } \left|\alpha _{n}\right|^{2} r^{2n} \qquad (0 < r < 1)

be bounded.

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Let the function φ\varphi be monotone and bounded on the interval [0,+∞)[0,+\infty ). Prove that for every convergent series ∑an\sum a_{n}

∑φ(εn)an⟶φ(+0)∑an as ε→+0. \sum \varphi (\varepsilon n) a_{n} \longrightarrow \varphi (+0) \sum a_{n} \quad \text{ as } \quad \varepsilon \rightarrow +0 .
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Let an∈R,bn>0,an=O(bn),∑bn=+∞a_{n} \in \mathbb {R}, b_{n}>0, a_{n}=O\left(b_{n}\right), \sum b_{n}=+\infty and let the series B(t)=∑n⩾0bntnB(t)=\sum_{n \geqslant 0} b_{n} t^{n} converge for ∣t∣<1\left|t\right|<1. Prove that the series A(t)=∑n⩾0antnA(t)=\sum_{n \geqslant 0} a_{n} t^{n} converges for ∣t∣<1\left|t\right|<1 and

lim‾⁡anbn⩽lim‾⁡t→1−0A(t)B(t)⩽lim‾⁡t→1−0A(t)B(t)⩽lim‾⁡anbn \varliminf \frac{a_{n}}{b_{n}} \leqslant \varliminf _{t \rightarrow 1-0} \frac{A(t)}{B(t)} \leqslant \varlimsup _{t \rightarrow 1-0} \frac{A(t)}{B(t)} \leqslant \varlimsup \frac{a_{n}}{b_{n}}

(compare with problem II.2.2). In particular, if an∼cbn,c≠0a_{n} \sim c b_{n}, c \neq 0, then A(t)∼cB(t)A(t) \sim c B(t) as t→1−0t \rightarrow 1-0; if an=o(bn)a_{n}=o\left(b_{n}\right), then A(t)=o(B(t))A(t)=o(B(t)) as t→1−0t \rightarrow 1-0.

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If the partial sums SnS_{n} of the series ∑n⩾0an\sum_{n \geqslant 0} a_{n} are such that the finite limit lim⁡1n(S1+…+Sn)=l\lim \frac{1}{n}\left(S_{1}+\ldots +S_{n}\right)=l exists, then the series A(t)=∑n⩾0antnA(t)=\sum_{n \geqslant 0} a_{n} t^{n} converges for ∣t∣<1\left|t\right|<1 and A(t)⟶t→1−0lA(t) \underset {t \rightarrow 1-0}{\longrightarrow } l (Frobenius's theorem).

Thus Abel's method of summing series (computing the limit lim⁡t→1−0∑n⩾0antn\lim_{t \rightarrow 1-0} \sum_{n \geqslant 0} a_{n} t^{n}) is stronger than the method of arithmetic means (Cesàro's method).

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Let bn⩾0,∑bn=+∞b_{n} \geqslant 0, \sum b_{n}=+\infty and let the series B(t)=∑n⩾0bntnB(t)=\sum_{n \geqslant 0} b_{n} t^{n} converge for ∣t∣<1\left|t\right|<1. If the sequence {an}n⩾0\left\{ a_{n}\right\}_{n \geqslant 0} is such that the finite limit

lim⁡a0+…+anb0+…+bn=l, \lim \frac{a_{0}+\ldots +a_{n}}{b_{0}+\ldots +b_{n}}=l,

exists, then the series A(t)=∑n⩾0antnA(t)=\sum_{n \geqslant 0} a_{n} t^{n} converges for ∣t∣<1\left|t\right|<1 and A(t)B(t)→t→1−0l\frac{A(t)}{B(t)} \xrightarrow [t \rightarrow 1-0]{ } l.

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Let the sequence {an}\left\{ a_{n}\right\} be such that an>−1a_{n}>-1 for n∈Nn \in \mathbb {N} and Π(1+an)=A∈(0,+∞)\Pi \left(1+a_{n}\right)=A \in (0,+\infty ). Prove that for ∣t∣<1\left|t\right|<1 the product A(t)=∏(1+antn)A(t)=\prod \left(1+a_{n} t^{n}\right) converges and, if ∑an2<+∞\sum a_{n}^{2}<+\infty, then A(t)→AA(t) \rightarrow A as t→1−0t \rightarrow 1-0.

Can the condition ∑an2<+∞\sum a_{n}^{2}<+\infty be dropped?

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Suppose the integral ∫0∞f(x)dx\int_{0}^{\infty } f(x) d x converges. Prove that

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(a)

for any ε>0\varepsilon >0 the integral ∫0∞e−εxf(x)dx\int_{0}^{\infty } e^{-\varepsilon x} f(x) d x converges;

(b)

∫0∞e−εxf(x)dx→∫0∞f(x)dx\int_{0}^{\infty } e^{-\varepsilon x} f(x) d x \rightarrow \int_{0}^{\infty } f(x) d x as ε→+0\varepsilon \rightarrow +0 (compare with Abel's theorem — Problem IV.5.8).

Compute the integral ∫0∞sin⁡xxdx\int_{0}^{\infty } \frac{\sin x}{x} d x.

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Let the functions

φ0(t),φ1(t),φ2(t),…,φn(t),… \varphi _0(t), \varphi _1(t), \varphi _2(t), \ldots , \varphi _n(t), \ldots

be non-negative in the interval 0<t<10<t<1 and assume that

φ0(t)+φ1(t)+φ2(t)+⋯+φn(t)+⋯=1 \varphi _0(t)+\varphi _1(t)+\varphi _2(t)+\cdots +\varphi _n(t)+\cdots = 1

holds for all 0<t<10<t<1. The sequence s0,s1,s2,…,sn,…s_0, s_1, s_2, \ldots , s_n, \ldots is supposed to be bounded. Construct the function

Φ(t)=s0φ0(t)+s1φ1(t)+s2φ2(t)+⋯+snφn(t)+⋯ . \Phi (t) = s_0 \varphi _0(t) + s_1 \varphi _1(t) + s_2 \varphi _2(t) + \cdots + s_n \varphi _n(t) + \cdots .

The range of Φ(t)\Phi (t) will fall into the interval between the upper and lower bounds of the sequence s0,s1,s2,…,sn,…s_0, s_1, s_2, \ldots , s_n, \ldots

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(continued). Show that

lim⁡t→1−0(s0φ0(t)+s1φ1(t)+s2φ2(t)+⋯+snφn(t)+⋯)=s \lim _{t \rightarrow 1-0} \big( s_0 \varphi _0(t) + s_1 \varphi _1(t) + s_2 \varphi _2(t) + \cdots + s_n \varphi _n(t) + \cdots \big) = s

holds for every convergent sequence s0,s1,s2,…s_0, s_1, s_2, \ldots for which lim⁡n→∞sn=s\lim_{n \rightarrow \infty } s_n = s if and only if for each fixed ν\nu

lim⁡t→1−0φν(t)=0. \lim _{t \rightarrow 1-0} \varphi _{\nu }(t) = 0 .
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The infinite sequences

a0,a1,a2,…,an,…;b0,b1,b2,…,bn,… a_0, a_1, a_2, \ldots , a_n, \ldots ; \qquad b_0, b_1, b_2, \ldots , b_n, \ldots

satisfy the following three conditions:

bn>0,n=0,1,2,…;b0+b1t+b2t2+⋯{is convergent for ∣t∣<1,divergent for t=1; b_n>0, \quad n=0,1,2,\ldots ; \qquad b_0+b_1t+b_2t^2+\cdots \begin{cases} \text{is convergent for } \left|t\right|<1, \\ \text{divergent for } t=1 ; \end{cases} lim⁡n→∞anbn=s. \lim _{n \rightarrow \infty } \frac{a_n}{b_n} = s .

Then

a0+a1t+a2t2+⋯+antn+⋯ converges for ∣t∣<1 a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots \text{ converges for } \left|t\right|<1

and

lim⁡t→1−0a0+a1t+a2t2+⋯+antn+⋯b0+b1t+b2t2+⋯+bntn+⋯=s. \lim _{t \rightarrow 1-0} \frac{a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots }{b_0+b_1t+b_2t^2+\cdots +b_nt^n+\cdots } = s .
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Remark.
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This proposition is due to E. Cesàro. Several applications will be given in the sequel.

Set

sn=a0+a1+a2+⋯+an,n=0,1,2,3,… s_n = a_0+a_1+a_2+\cdots +a_n, \qquad n=0,1,2,3,\ldots

If

lim⁡n→∞s0+s1+s2+⋯+snn+1=s \lim _{n \rightarrow \infty } \frac{s_0+s_1+s_2+\cdots +s_n}{n+1} = s

exists, then

lim⁡t→1−0(a0+a1t+a2t2+⋯+antn+⋯ )=s. \lim _{t \rightarrow 1-0} (a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots ) = s .
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Remark.
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This proposition goes beyond problem 86 only if the series a0+a1+a2+⋯+an+⋯a_0+a_1+a_2+\cdots +a_n+\cdots diverges (problem 67). G. Frobenius: J. reine angew. Math. Vol. 89, pp. 262-264 (1880).

If the following conditions are satisfied:

bn>0,∑n=0∞bn divergent,lim⁡n→∞a0+a1+a2+⋯+anb0+b1+b2+⋯+bn=s, b_n>0, \qquad \sum _{n=0}^{\infty } b_n \text{ divergent}, \qquad \lim _{n \rightarrow \infty } \frac{a_0+a_1+a_2+\cdots +a_n}{b_0+b_1+b_2+\cdots +b_n} = s ,

then

lim⁡t→1−0a0+a1t+a2t2+⋯+antn+⋯b0+b1t+b2t2+⋯+bntn+⋯=s, \lim _{t \rightarrow 1-0} \frac{a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots }{b_0+b_1t+b_2t^2+\cdots +b_nt^n+\cdots } = s ,

provided that the series in the denominator converges for ∣t∣<1\left|t\right|<1.

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The following limit exists and is positive provided α\alpha is positive:

lim⁡t→1−0(1−t)α(1α−1t+2α−1t2+3α−1t3+⋯+nα−1tn+⋯). \lim _{t \rightarrow 1-0} (1-t)^{\alpha } \big( 1^{\alpha -1}t + 2^{\alpha -1}t^2 + 3^{\alpha -1}t^3 + \cdots + n^{\alpha -1}t^n + \cdots \big) .
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Let AnA_n and BnB_n be the numerator and denominator, respectively, of the nn-th convergent of the infinite continued fraction

a1+a3+a5+⋯and soAnBn=a1+a3+a5+⋯+a2n−3,a>0. \cfrac {a}{1} + \cfrac {a}{3} + \cfrac {a}{5} + \cdots \quad \text{and so} \quad \frac{A_n}{B_n} = \cfrac {a}{1} + \cfrac {a}{3} + \cfrac {a}{5} + \cdots + \cfrac {a}{2n-3} , \quad a>0 .

Assume that this continued fraction converges. Find its value applying problem 85 and using the series

F(x)=∑n=0∞Ann!xn,G(x)=∑n=0∞Bnn!xn. F(x) = \sum _{n=0}^{\infty } \frac{A_n}{n!} x^n , \qquad G(x) = \sum _{n=0}^{\infty } \frac{B_n}{n!} x^n .
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Remark.
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F(x)F(x) and G(x)G(x) satisfy a linear homogeneous differential equation of the second order by virtue of the recursion formulas for AnA_n and BnB_n. Cf.O. Perron: Die Lehre von den Kettenbrüchen. Leipzig: B. G. Teubner 1913, p. 353, formula (24); R. G. Archibald: An Introduction to the Theory of Numbers. Columbus/Ohio: Charles E. Merrill Publishing 1970, p. 176.

Let σ>0\sigma >0. If the series

a11−σ+a22−σ+a33−σ+⋯+ann−σ+⋯ a_1 1^{-\sigma } + a_2 2^{-\sigma } + a_3 3^{-\sigma } + \cdots + a_n n^{-\sigma } + \cdots

is convergent then we have

lim⁡t→1−0(1−t)σ(a1t+a2t2+a3t3+⋯+antn+⋯ )=0[see problem 75]. \lim _{t \rightarrow 1-0} (1-t)^{\sigma } (a_1 t + a_2 t^2 + a_3 t^3 + \cdots + a_n t^n + \cdots ) = 0 \qquad \text{[see problem 75].}
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Show that

lim⁡t→1−01−t∑n=1∞(tn2−2t2n2) \lim _{t \rightarrow 1-0} \sqrt{1-t} \sum _{n=1}^{\infty } \big( t^{n^2} - 2t^{2n^2} \big)

exists and is negative.

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The two given sequences

a0,a1,a2,…,an,…;b0,b1,b2,…,bn,… a_0, a_1, a_2, \ldots , a_n, \ldots ; \qquad b_0, b_1, b_2, \ldots , b_n, \ldots

satisfy the conditions

bn>0;∑n=0∞bntn converges for all values of t;lim⁡n→∞anbn=s. b_n>0; \qquad \sum _{n=0}^{\infty } b_n t^n \text{ converges for all values of } t ; \qquad \lim _{n \rightarrow \infty } \frac{a_n}{b_n} = s .

Then a0+a1t+a2t2+⋯+antn+⋯a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots converges too for all values of tt and in addition

lim⁡t→∞a0+a1t+a2t2+⋯+antn+⋯b0+b1t+b2t2+⋯+bntn+⋯=s.(Cf. Problem 72 of Part IV.) \lim _{t \rightarrow \infty } \frac{a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots }{b_0+b_1t+b_2t^2+\cdots +b_nt^n+\cdots } = s . \qquad \text{(Cf.\ Problem 72 of Part IV.)}
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If lim⁡n→∞sn=s\lim_{n \rightarrow \infty } s_n = s exists then

lim⁡t→∞(s0+s1t1!+s2t22!+⋯+sntnn!+⋯ )e−t=s. \lim _{t \rightarrow \infty } \left( s_0 + s_1 \frac{t}{1!} + s_2 \frac{t^2}{2!} + \cdots + s_n \frac{t^n}{n!} + \cdots \right) e^{-t} = s .
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Assume that the sum

a0+a1+a2+⋯+an+⋯=s a_0+a_1+a_2+\cdots +a_n+\cdots = s

exists. Define

g(t)=a0+a1t1!+a2t22!+⋯+antnn!+⋯ . g(t) = a_0 + a_1 \frac{t}{1!} + a_2 \frac{t^2}{2!} + \cdots + a_n \frac{t^n}{n!} + \cdots .

Then

∫0∞e−tg(t) dt=s. \int _0^{\infty } e^{-t} g(t) \, dt = s .
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The Γ\Gamma-function is defined for α>0\alpha > 0 (or ℜα>0\Re \alpha > 0) by the integral

Γ(α)=∫0∞e−xxα−1 dx. \Gamma (\alpha ) = \int _{0}^{\infty } e^{-x} x^{\alpha -1} \, dx .

Using Problem 89 of Part I prove that

Γ(α)=lim⁡n→∞nα−1n!α(α+1)⋯(α+n−1),α>0. \Gamma (\alpha ) = \lim _{n \to \infty } \frac{n^{\alpha -1} n!}{\alpha (\alpha +1) \cdots (\alpha +n-1)}, \qquad \alpha > 0 .
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lim⁡t→1−0(1−t)(t1+t+t21+t2+t31+t3+⋯+tn1+tn+⋯ )=log⁡2. \lim _{t \to 1-0} (1-t) \left(\frac{t}{1+t} + \frac{t^{2}}{1+t^{2}} + \frac{t^{3}}{1+t^{3}} + \cdots + \frac{t^{n}}{1+t^{n}} + \cdots \right) = \log 2 .
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lim⁡t→1−0(1−t)2(t1−t+2t21−t2+3t31−t3+⋯+ntn1−tn+⋯ )=π26. \lim _{t \to 1-0} (1-t)^{2} \left(\frac{t}{1-t} + 2 \frac{t^{2}}{1-t^{2}} + 3 \frac{t^{3}}{1-t^{3}} + \cdots + n \frac{t^{n}}{1-t^{n}} + \cdots \right) = \frac{\pi ^{2}}{6} .
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We have

lim⁡ε→+0ε∫0∞e−εtf(t) dt=lim⁡t→∞f(t), \lim _{\varepsilon \to +0} \varepsilon \int _{0}^{\infty } e^{-\varepsilon t} f(t) \, dt = \lim _{t \to \infty } f(t),

provided that the integral on the left and the limit on the right hand side exist.

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If the Dirichlet series

a11−s+a22−s+a33−s+⋯+ann−s+⋯ a_{1}1^{-s} + a_{2}2^{-s} + a_{3}3^{-s} + \cdots + a_{n}n^{-s} + \cdots

converges for s=σ+iτs = \sigma + i\tau, σ,τ\sigma , \tau real, σ>0\sigma > 0, then

lim⁡t→1−0(1−t)σ(a1t+a2t2+a3t3+⋯+antn+⋯ )=0.[Problem 92 of Part I.] \lim _{t\to 1-0} (1-t)^{\sigma }(a_{1}t + a_{2}t^{2} + a_{3}t^{3} + \cdots + a_{n}t^{n} + \cdots ) = 0. \quad [\text{Problem 92 of Part I.}]
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(Generalization of Problem 94 of Part I.) Let

b0+b1z+b2z2+⋯+bnzn+⋯=f(z) b_{0} + b_{1} z + b_{2} z^{2} + \cdots + b_{n} z^{n} + \cdots = f(z)

be an everywhere convergent power series. From the three conditions

bn>0,n=0,1,2,…;lim⁡r→+∞r−βlog⁡f(r)=b;lim⁡n→∞annkb=s b_{n} > 0, \quad n = 0, 1, 2, \ldots ; \qquad \lim _{r \to +\infty } r^{-\beta } \log f(r) = b; \qquad \lim _{n \to \infty } \frac{a_{n}}{n^{k} b} = s

(β,b\beta , b positive, kk and ss arbitrary real constants) it follows that

lim⁡r→+∞a0+a1r+a2r2+⋯+anrn+⋯(βbrβ)kf(r)=s. \lim _{r \to +\infty } \frac{a_{0} + a_{1} r + a_{2} r^{2} + \cdots + a_{n} r^{n} + \cdots }{(\beta b r^{\beta })^{k} f(r)} = s.
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