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Showing posts with label Series. Show all posts
Showing posts with label Series. Show all posts

Monday, December 5, 2016

Pseudo sum

Let $\alpha, \beta $ be positive irrational numbers such that $\displaystyle \frac{1}{\alpha} + \frac{1}{\beta}=1$. Evaluate the (pseudo) sum:


$$\sum_{n=1}^{\infty} \left(\frac{1}{\lfloor n\alpha\rfloor^2}+\frac{1}{\lfloor n\beta\rfloor^2}\right)$$

Solution

A double Putnam 2016 series

Evaluate the series:

$$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}$$

(Putnam 2016)
Solution

Tuesday, September 13, 2016

Double alternating sum

Let $\displaystyle c_n= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n}$. Evaluate the sum:

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

Solution

Monday, September 12, 2016

A beautiful double sum

Prove that:

$$ \sum_{n=1}^{\infty}\left [ \frac{1}{n} \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2n}} - \log \left ( 1 + \frac{1}{n} \right ) \right ]$$

Solution

Wednesday, September 7, 2016

A log integral

Prove that:

$$\int_{0}^{\pi/2}{x\log \left( 1-\cos x \right) \, {\rm d} x}=\frac{35}{16}\zeta \left( 3 \right)-\frac{\pi ^{2}}{8}\log 2-\pi\mathcal{G}$$

Solution

Monday, August 15, 2016

Series with trilogarithm

Let ${\rm Li}_3$ denote the trilogarithm function. Prove that:

$$\sum_{n=1}^{\infty} {\rm Li}_3 \left(e^{-2n \pi} \right)= \frac{7 \pi^3}{360} - \frac{\zeta(3)}{2}$$

(Seraphim Tsipelis)

Solution [by r9m]

Saturday, July 2, 2016

Alternating series with eta dirichlet

Let $\eta$ denote Dirichlet's eta function. Prove that

$$\frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}$$

Solution

$\sum \limits_{n=1}^{\infty} \left(\zeta(2n) -\beta(2n) \right)$

Prove that:

$$\sum_{n=1}^{\infty}\left( \zeta(2n)-\beta(2n) \right)=\frac{1}{2}+\frac{\ln 2}{2}$$

Solution

Thursday, June 30, 2016

A $\zeta(2n+1)$ series

Let $\zeta$ denote the Riemann Zeta function. Evaluate the series:

$$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)(2n+1)}$$

(Serafim Tsipelis, Anastasios Kotronis)
Solution

Tuesday, June 28, 2016

Diriclet series

Let $\sigma(n)$ be the divisor function that is $ \displaystyle {\rm \sigma(n)=\sum \limits_{d \mid n} d}$. Prove that for $s \in \mathbb{R} \mid s >2$ it holds that:

$$\sum_{n=1}^{\infty} \frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$

where $\zeta$ is the Riemann zeta function.

Solution

Series

Let $a>\frac{1}{4}$. Prove that:

$$\sum_{n=1}^{\infty} \frac{1}{n^2-n+a} = \frac{\pi}{\sqrt{4a-1}} \frac{e^{\pi \sqrt{4a-1}}-1}{e^{\pi \sqrt{4a-1}}+1}$$

Solution

Sunday, June 26, 2016

Double alternating sum

Evaluate the series

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n} \frac{m}{m+n}$$

Solution

Saturday, June 18, 2016

Saturday, June 4, 2016

Double lattice sums

Prove that

$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \log 2}{8}$$

Solution

Sunday, May 15, 2016

A zeta series

Evaluate the series:

$$\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta(3n)}{n}$$

Solution

A weird infinite product

Evaluate the following product

$$P =\left({\frac{2}{1}}\right)^{1/8}\cdot\left({\frac{{2\cdot 2}}{{1\cdot 3}}}\right)^{3/16}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 4}}{{1\cdot 3\cdot 3\cdot 3}}}\right)^{6/32}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 2\cdot 4\cdot 4\cdot 4\cdot 4}}{{1\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 5}}}\right)^{10/64}\cdots$$

Solution

Saturday, May 7, 2016

Sunday, May 1, 2016

Parametric integral with logarithm

Let $n \in \mathbb{N} \setminus \{1\}$. Prove that:

$$\int_{0}^{\infty} \frac{n^2 x^n \log x}{1+x^{2n}} \, {\rm d}x = \frac{\pi^2}{4} \frac{\sin \left ( \frac{\pi}{2n} \right )}{\cos^2 \left ( \frac{\pi}{2n} \right )}$$

Solution

Thursday, April 28, 2016

On a relation between Euler sums

Prove that:

$$\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} = \frac{\pi \log 2}{8} - \frac{\mathcal{G}}{2}$$

Solution

Sunday, March 27, 2016

Double series

Evaluate the double series:

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m^2n + mn^2 + 2mn}$$

Solution