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Peculiar alternating series

July 12, 2022 / Leave a comment

Prove that

$\sum_{n=1}^{\infty} \frac{1}{n} \left ( \frac{1}{(-1)^n - 5} \right )^n = \frac{1}{2} \ln \frac{16}{21}$

Solution

We simply note that

$\begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \left ( \frac{1}{(-1)^n - 5} \right )^n &= -\sum_{n=1}^{\infty} \frac{1}{2n-1} \left ( \frac{1}{6} \right )^{2n-1} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \left ( \frac{1}{16} \right )^n \\ &= \frac{1}{2} \ln \frac{1 - \frac{1}{6}}{1 + \frac{1}{6}} - \frac{1}{2} \ln \left ( 1 - \frac{1}{16} \right ) \\ &=\frac{1}{2} \ln \frac{16}{21} \end{align*}$

Hyperbolic series

June 8, 2022 / Leave a comment

Evaluate the series

$\mathcal{S} = \sum_{n=1}^{\infty} \frac{1}{\sinh 2^n}$

Solution

We have successively

$\begin{align*} \sum_{n=1}^{\infty} \frac{1}{\sinh 2^n} &= \sum_{n=1}^{\infty} \frac{2}{e^{2^n} - e^{-2^n}} \\ &= \sum_{n=1}^{\infty} \frac{2}{e^{2n} \left ( 1 - e^{-2 \cdot 2^n} \right )} \\ &= 2 \sum_{n=1}^{\infty} e^{-2^n} \sum_{m=0}^{\infty} e^{-2m \cdot 2^n }\\ &=2 \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} e^{-\left ( 2m+1 \right ) 2^n} \\ &=2 \sum_{j=1}^{\infty} e^{-2j} \\ &= \frac{2}{e^2-1} \end{align*}$

since by the unique factorization theorem, any positive even integer $2j$ can be written in a unique way as $(2m+1)2^n$ with $m \in \mathbb{N}$ and $n \in \mathbb{N}^+$ .

A logarithmic integral

May 14, 2022 / Leave a comment

Let $\alpha \in \mathbb{R}$ . Prove that:

$\int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2}\, \mathrm{d}x = \frac{\pi}{\sqrt{2}} \cdot \frac{\alpha}{\sqrt{1+ \sqrt{1+\alpha^2}}}$

Solution

Using the known Fourier series $\displaystyle \sum_{n=-\infty}^{\infty} \frac{1}{\left( x+2\pi n \right)^2}=\frac{1}{4} \csc^2 \frac{x}{2}$ as well as the known generating function of the Catalan numbers

$\sum_{n=0}^{\infty} \mathcal{C}_n x^n = \frac{1- \sqrt{1-4x}}{2x} \Rightarrow \sum_{n=0}^{\infty} \mathcal{C}_{2n} x^{2n} = \frac{\sqrt{1+4x} - \sqrt{1-4x}}{4x}$

Then, we have successively:

$\begin{align*} \int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2} \, \mathrm{d}x &=\frac{1}{2} \int_{-\infty}^{\infty}\frac{\arctan \left(\alpha \sin^{2} x\right)}{x^{2}}\, \mathrm{d}x\\ &=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan \left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}} \, \mathrm{d}x\\ &=\frac{1}{8}\int_{0}^{2\pi} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}\frac{x}{2} \, \mathrm{d}x\\ &=\frac{1}{4}\int_{0}^{\pi/2}\arctan \left(\alpha \sin^{2} x \right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right) \, \mathrm{d} x\\ &=\int_{0}^{\pi/2} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}x\, \mathrm{d} x\\ &=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{2n+1} \alpha^{2n+1} \int_{0}^{\pi/2}\sin^{4n}x \, \mathrm{d}x\\ &=\pi \alpha \sum_{n=0}^{\infty} \mathcal{C}_{2n}\left(-\frac{\alpha^{2}}{16}\right)^{n} \\ &=\frac{\pi}{i}\left(\sqrt{1+i \alpha}-\sqrt{1-i\alpha}\right)\\ &=\frac{\pi}{\sqrt{2}}\frac{\alpha}{\sqrt{1+\sqrt{1+\alpha^{2}}}} \end{align*}$

A Gamma summation

May 11, 2022 / Leave a comment

Let $a \notin \mathbb{Z}$ and $a > \frac{1}{2}$ . Prove that

$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$

Solution

Consider the function

$f(x) = \left\{\begin{matrix} \cos^{2a-2} \pi x & , & \left | x \right |< \frac{1}{2} \\ 0 & , & \text{elsewhere} \end{matrix}\right.$

We easily evaluate that

$\hat{f}(y)= \int_{-\infty}^{\infty} f(x) e^{-2i \pi x y} \, \mathrm{d}x = \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+y) \Gamma(a-y)}$

Hence it follows from Poisson summation formula that

$\begin{align*} 1 &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) \\ &= \sum_{n=-\infty}^{\infty} \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+n) \Gamma(a-n)} \end{align*}$

Thus,

$\begin{align*} \frac{1}{2^{2-2a} \Gamma(2a-1)} &= \sum_{n=-\infty}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &=\sum_{n=-\infty}^{-1} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(\alpha)} + \\ &\quad \quad \quad + \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &= 2 \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(a)} \\ &= 2 \sum_{n=0}^{\infty}\frac{1}{\Gamma(a+n) \Gamma(a-n)} - \frac{1}{\Gamma^2(a)} \end{align*}$

The result follows.

Factorial series

January 1, 2022 / Leave a comment

Prove that

$\sum_{n=0}^{\infty} \frac{(4n)!}{(4n+4)!} = \frac{\ln 2}{4} - \frac{\pi}{24}$

Solution

We have successively:

$\begin{align*} \sum_{n =0}^{\infty} \frac{(4n)!}{(4n+4)!} & =\sum_{n=0}^{\infty} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\ &= \frac{1}{6} \sum_{n=0}^{\infty} \mathrm{B}(4n+1,4) \\ & = \frac{1}{6}\sum_{n=0}^{\infty} \int_{0}^{1}x^{4n}(1-x)^3\, \mathrm{d}x \\ & = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x+1}{x^2+1} \right )\, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x}{x^2+1} - \frac{1}{x^2+1} \right )\, \mathrm{d}x \\ & = \frac{\ln 2}{4}-\frac{\pi}{24} \end{align*}$

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Peculiar alternating series

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Joy of Mathematics

Tag Archives: Series

Recent Posts

Peculiar alternating series

Hyperbolic series

A logarithmic integral

A Gamma summation

Factorial series

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Donate to Tolaso Network

General

Associated Sites

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