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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:

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

The composition is a metric

December 18, 2021 / Leave a comment

Let $d$ be a metric , $f$ be a strictly increasing function and concave on $[0, +\infty)$ such that $f(0)=0$ . Prove that $f \circ d$ is a metric.

Symmetry of Euler sums

December 7, 2021 / Leave a comment

Let $\zeta$ denote the Riemann zeta function. Prove that

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

where $\mathcal{H}_n^{(s)}$ is the generalized harmonic number of order $s$ .

Solution

We have successively

$\begin{align*} \zeta^2(k) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{n^km^k} \\&= 2\sum_{1\leq m\leq n} \frac{1}{(nm)^k} - \sum_{m=1}^\infty\frac{1}{(n^k)^2}\\ &=2\sum_{n=1}^\infty \frac{1}{n^k}\sum_{m=1}^n \frac{1}{m^k} - \zeta(2k)\\ &=2\sum_{n=1}^\infty \frac{\mathcal{H}_n^{(k)}}{n^k} - \zeta(2k) \end{align*}$

A limit

December 5, 2021 / Leave a comment

Let $f:(0, +\infty) \rightarrow \mathbb{R}$ . If the line $y=\lambda x + \beta \; , \; \lambda>0$ is an oblique asymptote of $\mathcal{C}_f$ at $+\infty$ then evaluate the limit

$\ell = \lim_{x \rightarrow +\infty} \left( f(x) \ln (x+1) - f(x+1) \ln x \right)$

An integral

November 27, 2021 / Leave a comment

Let $n \in \mathbb{Z}_{\geq 0}$ . Evaluate the integral

$\mathcal{J}_n = \int_0^\pi \frac{1- \cos nx}{1-\cos x}\, \mathrm{d}x$

Solution

We note that

$\begin{align*} \frac{\mathcal{J}_{n+1}+\mathcal{J}_{n-1}}{2} &=\int_{0}^{\pi}\frac{2-\cos (n+1)x-\cos (n-1)x}{2(1-\cos x)}\, \mathrm{d}x \\ &= \int_{0}^{\pi}\frac{1-\cos n x \cos x}{1-\cos x} \, \mathrm{d}x \\ &=\int_{0}^{\pi}\frac{\left ( 1-\cos n x \right )+\cos n x(1-\cos x)}{1-\cos x}\, \mathrm{d}x \\ &=\mathcal{J}_n+\int_{0}^{\pi} \cos n x\, \mathrm{d}x \\ &=\mathcal{J}_n \end{align*}$

This shows that we have a geometric progression. Since $\mathcal{J}_0 = 0$ , $\mathcal{J}_1 = \pi$ it follows that $\mathcal{J}_n = n \pi$ .

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