Section 12.4

Miscellaneous Problems

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Problem 5126

Prove Catalan's identity

1−12+13−14+…−12n=1n+1+1n+2+…+12n 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots -\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{2n}

and, using it, show that ∑n=1∞(−1)n−1n=ln⁡2\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n} = \ln 2.

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Problem 5127

Prove the equalities:

1+12+13+…+1n=C+ln⁡n+εn; 1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{n} = C+\ln n+\varepsilon _{n}; 1+13+15+…+12n−1=C2+ln⁡4n+ηn, 1+\frac{1}{3}+\frac{1}{5}+\ldots +\frac{1}{2n-1} = \frac{C}{2}+\ln \sqrt{4n}+\eta _{n},

where CC is Euler's constant, and εn\varepsilon_{n} and ηn→0\eta_{n} \rightarrow 0 as n→∞n \rightarrow \infty.

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Problem 5128

Prove the equalities:

1+13−12+15+17−14+19+111−16+…=32ln⁡2; 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\ldots = \frac{3}{2}\ln 2; 1−12−14+13−16−18+15−…=12ln⁡2; 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\ldots = \frac{1}{2}\ln 2; 1−12−14−16−18+13−110−112−114−116+15−…=0. 1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{3}-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac{1}{5}-\ldots = 0.
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Problem 5129

Let the series ∑un\sum u_{n} be obtained from the series 1−12+13−14+15−…1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots by a rearrangement of terms such that the terms of each sign decrease in absolute value, and among the first nn terms the ratio of the number of positive terms to the number of negative terms tends to σ\sigma as n→∞n \rightarrow \infty. Prove the equality: ∑ν=1∞uν=ln⁡4σ\sum_{\nu =1}^{\infty } u_{\nu } = \ln \sqrt{4\sigma }.

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Problem 5130

Prove the equalities:

1+15+19+…+14n−3=C4+π8+ln⁡8n4+εn;εn→0 as n→∞. 1+\frac{1}{5}+\frac{1}{9}+\ldots +\frac{1}{4n-3} = \frac{C}{4}+\frac{\pi }{8}+\ln \sqrt[4]{8n}+\varepsilon _{n}; \quad \varepsilon _{n} \rightarrow 0 \text{ as } n \rightarrow \infty . 13+17+111+…+14n−1=C4−π8+ln⁡8n4+δn;δn→0 as n→∞. \frac{1}{3}+\frac{1}{7}+\frac{1}{11}+\ldots +\frac{1}{4n-1} = \frac{C}{4}-\frac{\pi }{8}+\ln \sqrt[4]{8n}+\delta _{n}; \quad \delta _{n} \rightarrow 0 \text{ as } n \rightarrow \infty .
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Problem 5131

Prove the equality 1+15−13+19+113−17+…=π4+14ln⁡21+\frac{1}{5}-\frac{1}{3}+\frac{1}{9}+\frac{1}{13}-\frac{1}{7}+\ldots = \frac{\pi }{4}+\frac{1}{4}\ln 2.

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Problem 5132

Prove the equality 1−13−17+15−111−115+19−…=π4−14ln⁡21-\frac{1}{3}-\frac{1}{7}+\frac{1}{5}-\frac{1}{11}-\frac{1}{15}+\frac{1}{9}-\ldots = \frac{\pi }{4}-\frac{1}{4}\ln 2.

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Problem 5133

From the harmonic series 1+12+13+14+15+16+…1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\ldots, remove the terms whose denominator contains the digit 99. Prove that the resulting series converges, and that its sum is less than 8080.

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Problem 5134

Prove the equality s2+s3+s4+…=1s_{2}+s_{3}+s_{4}+\ldots = 1; sν=∑n=2∞1nνs_{\nu } = \sum_{n=2}^{\infty } \frac{1}{n^{\nu }}.

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Problem 5135

Prove the equality s3+s5+s7+…=14s_{3}+s_{5}+s_{7}+\ldots = \frac{1}{4}.

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Problem 5136

Prove the equality s2+12s4+13s6+…=ln⁡2s_{2}+\frac{1}{2}s_{4}+\frac{1}{3}s_{6}+\ldots = \ln 2.

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Problem 5137

Let ana_{n} be the numbers of the natural sequence of the form pqp^{q}, where pp and qq are integers greater than one. Prove Euler's equality: ∑n=1∞1an−1=1\sum_{n=1}^{\infty } \frac{1}{a_{n}-1} = 1.

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Problem 5138

Let ana_{n} be the numbers of the natural sequence of the form pqp^{q}, where pp and qq are integers greater than one. Prove Euler's equality: ∑n=1∞1an+1=π23−52\sum_{n=1}^{\infty } \frac{1}{a_{n}+1} = \frac{\pi^{2}}{3}-\frac{5}{2}.

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Problem 5139

Let ana_{n} be the numbers of the natural sequence of the form pqp^{q}, where pp and qq are integers greater than one, and let bmb_{m} be those of the numbers ana_{n} that are divisible by 22. Prove Euler's equality: ∑n=1∞1bm−1=ln⁡2\sum_{n=1}^{\infty } \frac{1}{b_{m}-1} = \ln 2.

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Problem 5140

Let ana_{n} be the numbers of the natural sequence of the form pqp^{q}, where pp and qq are integers greater than one, and let bmb_{m} be those of the numbers ana_{n} that are divisible by 22. Prove Euler's equality: ∑n=1∞1bm+1=π212+ln⁡2−1\sum_{n=1}^{\infty } \frac{1}{b_{m}+1} = \frac{\pi^{2}}{12}+\ln 2-1.

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Problem 5141

Prove the equality φ(x)+φ(1−x)=π26−ln⁡xln⁡(1−x)\varphi (x)+\varphi (1-x) = \frac{\pi^{2}}{6}-\ln x\ln (1-x); φ(x)=∑n=1∞xnn2\varphi (x) = \sum_{n=1}^{\infty } \frac{x^{n}}{n^{2}}.

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Problem 5142

Prove the equality φ(1−x)+φ(x−1x)=−12(ln⁡x)2\varphi (1-x)+\varphi \left(\frac{x-1}{x}\right) = -\frac{1}{2}(\ln x)^{2}, where φ(x)=∑n=1∞xnn2\varphi (x) = \sum_{n=1}^{\infty } \frac{x^{n}}{n^{2}} (the same function as in problem 5141).

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Problem 5143

Prove the equality ∑n=1∞1n2⋅2n=π212−12(ln⁡2)2\sum_{n=1}^{\infty } \frac{1}{n^{2}\cdot 2^{n}} = \frac{\pi^{2}}{12}-\frac{1}{2}(\ln 2)^{2}.

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Problem 5144

Prove the equality ∑n=1∞(−2)nsnn!=0\sum_{n=1}^{\infty } \frac{(-2)^{n}s_{n}}{n!} = 0; sn=13+23+33+…+n3s_{n} = 1^{3}+2^{3}+3^{3}+\ldots +n^{3}.

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Problem 5145

Prove the equality ∑n=0∞nmn!=e⋅pm\sum_{n=0}^{\infty } \frac{n^{m}}{n!} = e\cdot p_{m}, where the integers pmp_{m} are related by the equality:

pm+1=1+mp1+m(m−1)1⋅2p2+…+pm. p_{m+1} = 1+mp_{1}+\frac{m(m-1)}{1\cdot 2}p_{2}+\ldots +p_{m}.
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Problem 5146

Prove the equality

pσ+σ1pσ−1q+σ(σ−1)2!pσ−2q2+…+σ(σ−1)…(σ−n+1)n!pσ−nqn= p^{\sigma }+\frac{\sigma }{1}p^{\sigma -1}q+\frac{\sigma (\sigma -1)}{2!}p^{\sigma -2}q^{2}+\ldots +\frac{\sigma (\sigma -1)\ldots (\sigma -n+1)}{n!}p^{\sigma -n}q^{n} = =Γ(σ+1)n!Γ(σ−n)∫0pxσ−n−1(1−x)n dx; = \frac{\Gamma (\sigma +1)}{n!\Gamma (\sigma -n)}\int _{0}^{p} x^{\sigma -n-1}(1-x)^{n}\, dx;

here σ>n\sigma > n, 0<p<10 < p < 1, p+q=1p+q = 1.

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Problem 5147

Prove that the sum of the series

a1a2+x+a1a2+x⋅a2a3+x+a1a2+x⋅a2a3+x⋅a3a4+x+… \frac{a_{1}}{a_{2}+x}+\frac{a_{1}}{a_{2}+x}\cdot \frac{a_{2}}{a_{3}+x}+\frac{a_{1}}{a_{2}+x}\cdot \frac{a_{2}}{a_{3}+x}\cdot \frac{a_{3}}{a_{4}+x}+\ldots

equals a1x\frac{a_{1}}{x}, if the series ∑1an\sum \frac{1}{a_{n}} is divergent, and equals

a1x(1−∏ν=2∞aνaν+x), \frac{a_{1}}{x}\left(1-\prod _{\nu =2}^{\infty } \frac{a_{\nu }}{a_{\nu }+x}\right),

if the series ∑1an\sum \frac{1}{a_{n}} is convergent. The numbers x,a1,a2,a3,…x, a_{1}, a_{2}, a_{3}, \ldots are positive.

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Problem 5148

Prove the equality ∑n=2∞n(1+2x)(1+3x)…(1+nx)=1x\sum_{n=2}^{\infty } \frac{n}{(1+2x)(1+3x)\ldots (1+nx)} = \frac{1}{x}; x>0x > 0.

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Problem 5149

Prove the equality ∑n=2∞((n−1)!)2(x2+22)(x2+32)…(x2+n2)=1x2(1−πx(x2+1)sh⁡πx)\sum_{n=2}^{\infty } \frac{((n-1)!)^{2}}{(x^{2}+2^{2})(x^{2}+3^{2})\ldots (x^{2}+n^{2})} = \frac{1}{x^{2}}\left(1-\frac{\pi x(x^{2}+1)}{\operatorname {sh}\pi x}\right).

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Problem 5150

Find the sum of the series ∑n=1∞1a2+n2\sum_{n=1}^{\infty } \frac{1}{a^{2}+n^{2}}.

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Problem 5151

Find the sum of the series ∑n=1∞1(a2+n2)2\sum_{n=1}^{\infty } \frac{1}{(a^{2}+n^{2})^{2}}.

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Problem 5152

Find the sum of the series ∑−∞∞1x4+n4\sum_{-\infty }^{\infty } \frac{1}{x^{4}+n^{4}}.

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Problem 5153

Prove the following statement: lim⁡x→∞∑n=1∞x2x2+n2=π2\lim_{x \rightarrow \infty } \sum_{n=1}^{\infty } \frac{x^{2}}{x^{2}+n^{2}} = \frac{\pi }{2}.

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Problem 5154

Prove the following statement: if un>0u_{n} > 0, un−un+1>0u_{n}-u_{n+1} > 0, un−2un+1+un+2>0u_{n}-2u_{n+1}+u_{n+2} > 0, then

u02<u0−u1+u2−u3+u4−…<u0−12u1. \frac{u_{0}}{2} < u_{0}-u_{1}+u_{2}-u_{3}+u_{4}-\ldots < u_{0}-\frac{1}{2}u_{1}.
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Problem 5155

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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Problem 5156

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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Problem 5157

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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Problem 5158

Prove the following statement: lim⁡x→1−0(1−x)p∑n=1∞np−1xn=Γ(p)\lim_{x \rightarrow 1-0} (1-x)^{p}\sum_{n=1}^{\infty } n^{p-1}x^{n} = \Gamma (p); p>0p > 0.

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Problem 5159

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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Problem 5160

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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Problem 5161

Prove the following statement: lim⁡x→1−01−x (x+x4+x9+…)=12π\lim_{x \rightarrow 1-0} \sqrt{1-x}\, (x+x^{4}+x^{9}+\ldots ) = \frac{1}{2}\sqrt{\pi }.

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Problem 5162

Prove Poisson's formula:

∑n=abf(n)=f(a)+f(b)2+∫abf(x) dx+2∑n=1∞∫abf(x)cos⁡2πnx dx. \sum _{n=a}^{b} f(n) = \frac{f(a)+f(b)}{2} + \int _{a}^{b} f(x)\, dx + 2\sum _{n=1}^{\infty } \int _{a}^{b} f(x)\cos 2\pi nx\, dx.
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Problem 5163

Prove the equality 12+∑n=1∞e−πn2x=1x(12+∑n=1∞e−πn2x)\frac{1}{2}+\sum_{n=1}^{\infty } e^{-\pi n^{2}x} = \frac{1}{\sqrt{x}}\left(\frac{1}{2}+\sum_{n=1}^{\infty } e^{-\frac{\pi n^{2}}{x}}\right).

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Problem 5164

Prove the following statement: lim⁡x→1−0(x−x4+x9−x16+…)=12\lim_{x \rightarrow 1-0} (x-x^{4}+x^{9}-x^{16}+\ldots ) = \frac{1}{2}.

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Problem 5165

Prove the equality ∑n=−∞∞e−πa(n+x)2=1a∑n=−∞∞e−πn2acos⁡2πnx\sum_{n=-\infty }^{\infty } e^{-\pi a(n+x)^{2}} = \frac{1}{\sqrt{a}}\sum_{n=-\infty }^{\infty } e^{-\frac{\pi n^{2}}{a}}\cos 2\pi nx.

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Problem 5166

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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Problem 5167

Prove the following statement: if R1R_{1} and R2R_{2} are the radii of convergence of the series ∑anxn\sum a_{n}x^{n} and ∑bnxn\sum b_{n}x^{n}, then the radius of convergence of the series ∑anbnxn\sum a_{n}b_{n}x^{n} is no less than R1R2R_{1}R_{2}.

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Problem 5168

Prove the following statement: for the convergence of a series ∑an\sum a_{n} with positive decreasing terms, the condition lim⁡n→∞nan=0\lim_{n \rightarrow \infty } na_{n} = 0 is necessary.

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Problem 5169

Prove the following statement: if f(x)<Mf(x) < M for 0<x<2π0 < x < 2\pi, then for its Fourier coefficients the inequalities

∣an∣<2M−a0;∣bn∣<2M−a0 \left|a_{n}\right| < 2M-a_{0}; \quad \left|b_{n}\right| < 2M-a_{0}

hold.

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Problem 5170

Prove the following statement: the series ∑cos⁡xnnσ\sum \frac{\cos x\sqrt{n}}{n^{\sigma }} and ∑sin⁡xnnσ\sum \frac{\sin x\sqrt{n}}{n^{\sigma }} for x≠0x \neq 0 converge for σ>12\sigma > \frac{1}{2} and diverge for σ<12\sigma < \frac{1}{2}.

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Problem 5171

Prove the following statement: the series ∑(−1)[n]nσ\sum \frac{(-1)^{\left[\sqrt{n}\right]}}{n^{\sigma }}, where [n]\left[\sqrt{n}\right] is the integer part of n\sqrt{n}, converges for σ>12\sigma > \frac{1}{2} and diverges for σ<12\sigma < \frac{1}{2}.

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Problem 5172

Using the equality 1a=∫01xa−1 dx\frac{1}{a} = \int_{0}^{1} x^{a-1}\, dx, a>0a > 0, prove the formula:

1m−nm+p+n(n−1)2!⋅1m+2p−…+(−1)n1m+np= \frac{1}{m}-\frac{n}{m+p}+\frac{n(n-1)}{2!}\cdot \frac{1}{m+2p}-\ldots +(-1)^{n}\frac{1}{m+np} = =n!pnm(m+p)…(m+np). = \frac{n!p^{n}}{m(m+p)\ldots (m+np)}.
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Problem 5173

Prove the identity:

(1+x)(1+qx)…(1+qn−1x)= (1+x)(1+qx)\ldots (1+q^{n-1}x) = =1+qn−1q−1x+(qn−1)(qn−q)(q−1)(q2−1)x2+(qn−1)(qn−q)(qn−q2)(q−1)(q2−1)(q3−1)x3+… = 1+\frac{q^{n}-1}{q-1}x+\frac{(q^{n}-1)(q^{n}-q)}{(q-1)(q^{2}-1)}x^{2}+\frac{(q^{n}-1)(q^{n}-q)(q^{n}-q^{2})}{(q-1)(q^{2}-1)(q^{3}-1)}x^{3}+\ldots
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Remark.
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When xx is replaced by qxqx, the left-hand side is multiplied by 1+qnx1+q^{n}x and divided by 1+x1+x.

Problem 5174

Prove the equality for ∣q∣<1\left|q\right| < 1:

(1+qx)(1+q2x)(1+q3x)…=1+q1−qx+ (1+qx)(1+q^{2}x)(1+q^{3}x)\ldots = 1+\frac{q}{1-q}x+ +q3(1−q)(1−q2)x2+q6(1−q)(1−q2)(1−q3)x3+… +\frac{q^{3}}{(1-q)(1-q^{2})}x^{2}+\frac{q^{6}}{(1-q)(1-q^{2})(1-q^{3})}x^{3}+\ldots
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Problem 5175

Prove the equality for ∣q∣<1\left|q\right| < 1:

(1+qx)(1+q3x)(1+q5x)…=1+q1−q2x+ (1+qx)(1+q^{3}x)(1+q^{5}x)\ldots = 1+\frac{q}{1-q^{2}}x+ +q4(1−q2)(1−q4)x2+q9(1−q2)(1−q4)(1−q6)x3+… +\frac{q^{4}}{(1-q^{2})(1-q^{4})}x^{2}+\frac{q^{9}}{(1-q^{2})(1-q^{4})(1-q^{6})}x^{3}+\ldots
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Problem 5176

Prove the equality for ∣q∣<1\left|q\right| < 1:

∏ν=0n−1(1+q2ν+1x)(1+q2ν+1x−1)=A0+∑ν=1∞Aν(xν+x−ν); \prod _{\nu =0}^{n-1} (1+q^{2\nu +1}x)(1+q^{2\nu +1}x^{-1}) = A_{0}+\sum _{\nu =1}^{\infty } A_{\nu }(x^{\nu }+x^{-\nu }); Aν=qν2⋅(1−q2n+2ν+2)(1−q2n+2ν+4)…(1−q4n)(1−q2)(1−q4)…(1−q2n−2ν). A_{\nu } = q^{\nu ^{2}} \cdot \frac{(1-q^{2n+2\nu +2})(1-q^{2n+2\nu +4})\ldots (1-q^{4n})}{(1-q^{2})(1-q^{4})\ldots (1-q^{2n-2\nu })}.
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Problem 5177

Using the preceding result (problem 5176), obtain the Jacobi identity:

1+∑ν=1∞qν2(xν+x−ν)=∏ν=0∞(1+q2ν+1x)(1+q2ν+1x−1)(1−q2ν). 1+\sum _{\nu =1}^{\infty } q^{\nu ^{2}}(x^{\nu }+x^{-\nu }) = \prod _{\nu =0}^{\infty } (1+q^{2\nu +1}x)(1+q^{2\nu +1}x^{-1})(1-q^{2\nu }).
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Problem 5178

By a change of variables from the preceding result (problem 5177), obtain Euler's identity:

∑n=−∞∞(−1)nq3n2−n2=∏n=1∞(1−qn). \sum _{n=-\infty }^{\infty } (-1)^{n} q^{\frac{3n^{2}-n}{2}} = \prod _{n=1}^{\infty } (1-q^{n}).
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Problem 5179

Prove the Jacobi identities (for ∣q∣<1\left|q\right| < 1):

∑n=−∞∞(−1)nq(n+12)2x2n+1=q0⋅q14∏n=1∞(1−q2nx2)(1−q2nx−2)(x−x−1); \sum _{n=-\infty }^{\infty } (-1)^{n} q^{\left(n+\frac{1}{2}\right)^{2}} x^{2n+1} = q_{0} \cdot q^{\frac{1}{4}} \prod _{n=1}^{\infty } (1-q^{2n}x^{2})(1-q^{2n}x^{-2})(x-x^{-1}); ∑n=−∞∞q(n+12)2x2n+1=q0⋅q14(x+x−1)⋅∏n=1∞(1+q2nx2)(1+q2nx−2); \sum _{n=-\infty }^{\infty } q^{\left(n+\frac{1}{2}\right)^{2}} x^{2n+1} = q_{0} \cdot q^{\frac{1}{4}} (x+x^{-1}) \cdot \prod _{n=1}^{\infty } (1+q^{2n}x^{2})(1+q^{2n}x^{-2}); ∑n=−∞∞(−1)nqn2x2n=q0⋅∏n=1∞(1−q2n−1x2)(1−q2n−1x−2), \sum _{n=-\infty }^{\infty } (-1)^{n} q^{n^{2}} x^{2n} = q_{0} \cdot \prod _{n=1}^{\infty } (1-q^{2n-1}x^{2})(1-q^{2n-1}x^{-2}),

where q0=∏n=1∞(1−q2n)q_{0} = \prod_{n=1}^{\infty } (1-q^{2n}).

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Problem 5180

The Jacobi theta functions are defined by the equalities (for ∣q∣<1\left|q\right| < 1):

ϑ1(v)=2∑n=0∞(−1)nq(n+12)2sin⁡(2n+1)πv; \vartheta _{1}(v) = 2\sum _{n=0}^{\infty } (-1)^{n} q^{\left(n+\frac{1}{2}\right)^{2}} \sin (2n+1)\pi v; ϑ2(v)=2∑n=0∞q(n+12)2cos⁡(2n+1)πv; \vartheta _{2}(v) = 2\sum _{n=0}^{\infty } q^{\left(n+\frac{1}{2}\right)^{2}} \cos (2n+1)\pi v; ϑ3(v)=1+2∑n=1∞qn2cos⁡2πnv; \vartheta _{3}(v) = 1+2\sum _{n=1}^{\infty } q^{n^{2}} \cos 2\pi nv; ϑ4(v)=1+2∑n=1∞(−1)nqn2cos⁡2πnv. \vartheta _{4}(v) = 1+2\sum _{n=1}^{\infty } (-1)^{n} q^{n^{2}} \cos 2\pi nv.

Prove the equalities (for ∣q∣<1\left|q\right| < 1):

ϑ1(v)=2q0⋅q14sin⁡πv⋅∏n=1∞(1−2q2ncos⁡2πv+q4n); \vartheta _{1}(v) = 2q_{0} \cdot q^{\frac{1}{4}} \sin \pi v \cdot \prod _{n=1}^{\infty } (1-2q^{2n}\cos 2\pi v+q^{4n}); ϑ2(v)=2q0⋅q14cos⁡πv⋅∏n=1∞(1+2q2ncos⁡2πv+q4n); \vartheta _{2}(v) = 2q_{0} \cdot q^{\frac{1}{4}} \cos \pi v \cdot \prod _{n=1}^{\infty } (1+2q^{2n}\cos 2\pi v+q^{4n}); ϑ3(v)=q0⋅∏n=1∞(1+2q2n−1cos⁡2πv+q4n−2); \vartheta _{3}(v) = q_{0} \cdot \prod _{n=1}^{\infty } (1+2q^{2n-1}\cos 2\pi v+q^{4n-2}); ϑ4(v)=q0⋅∏n=1∞(1−2q2n−1cos⁡2πv+q4n−2), \vartheta _{4}(v) = q_{0} \cdot \prod _{n=1}^{\infty } (1-2q^{2n-1}\cos 2\pi v+q^{4n-2}),

where q0q_{0} has the same value as before (see problem 5179): q0=∏n=1∞(1−q2n)q_{0} = \prod_{n=1}^{\infty } (1-q^{2n}).

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Problem 5181

Setting

A=∏n=1∞(1+q2n);B=∏n=1∞(1+q2n−1);C=∏n=1∞(1−q2n−1) A = \prod _{n=1}^{\infty } (1+q^{2n}); \quad B = \prod _{n=1}^{\infty } (1+q^{2n-1}); \quad C = \prod _{n=1}^{\infty } (1-q^{2n-1})

and f(q)=ABCf(q) = ABC, where ∣q∣<1\left|q\right| < 1, prove that f(q)=f(q2)f(q) = f(q^{2}) and, consequently, ABC=1ABC = 1.

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Problem 5182

Prove Jacobi's equality: ϑ1′=πϑ2ϑ3ϑ4\vartheta_{1}' = \pi \vartheta_{2}\vartheta_{3}\vartheta_{4}, where ϑ2,ϑ3,ϑ4\vartheta_{2}, \vartheta_{3}, \vartheta_{4}, ϑ1′\vartheta_{1}' are the values of the theta functions and of their derivative at v=0v = 0 (see problem 5180 for the definition of the theta functions).

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Problem 5183

Prove the equality N=a(1+β+1⋅32!β2+1⋅3⋅53!β3+…)\sqrt{N} = a\left(1+\beta +\frac{1\cdot 3}{2!}\beta^{2}+\frac{1\cdot 3\cdot 5}{3!}\beta^{3}+\ldots \right), where N=a2+bN = a^{2}+b; β=b2N\beta = \frac{b}{2N}; ∣b∣<N\left|b\right| < N.

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Problem 5184

Prove the equality N3=a(1+β+1⋅42!β2+1⋅4⋅73!β3+…)\sqrt[3]{N} = a\left(1+\beta +\frac{1\cdot 4}{2!}\beta^{2}+\frac{1\cdot 4\cdot 7}{3!}\beta^{3}+\ldots \right), where N=a3+bN = a^{3}+b; β=b3N\beta = \frac{b}{3N}; ∣b∣<N\left|b\right| < N.

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Problem 5185

Verify the equality 2=1,41421 35623 73095 04880 16885…\sqrt{2} = 1{,}41421\ 35623\ 73095\ 04880\ 16885\ldots

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Problem 5186

Verify the equality 23=1,25991 20498 94873 16476 72106…\sqrt[3]{2} = 1{,}25991\ 20498\ 94873\ 16476\ 72106\ldots

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Problem 5187

Verify Ramanujan's equality:

11001+110022+310033+4210044+5310055+…= \frac{1}{1001}+\frac{1}{1002^{2}}+\frac{3}{1003^{3}}+\frac{4^{2}}{1004^{4}}+\frac{5^{3}}{1005^{5}}+\ldots = =11000−10−440⋅1,0125… = \frac{1}{1000}-10^{-440}\cdot 1{,}0125\ldots
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