spiderifilms6e

## Answered question

2022-01-21

Prove that:
$\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{\left(2n+1\right)}^{2m+1}}=\frac{{\left(-1\right)}^{m}{E}_{2m}{\pi }^{2m+1}}{{4}^{m+1}\left(2m\right)!}$

### Answer & Explanation

nick1337

Expert2022-01-27Added 777 answers

Because it wasnt

star233

Skilled2022-01-27Added 403 answers

The Dirichlet beta function is defined as $\beta \left(2m+1\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{\left(2n+1{\right)}^{2m+1}}$ Then $\beta \left(1\right)=\frac{\pi }{4}$ and $\beta \left(2m+1\right)=-\sum _{k=1}^{m}\frac{\left(-{\pi }^{2}/4{\right)}^{k}}{\left(2k\right)!}\beta \left(2m-2k+1\right)$ (1) If we reindex recursion derived below, we get that the even Euler numbers are defined by ${E}_{0}=1$ and ${E}_{2m}=-\sum _{k=1}^{m}\left(\left(2m\right),\left(2k\right)\right){E}_{2m-2k}$ (2) then notice that (1) is the same as (2) if we set $\beta \left(2m+1\right)=\frac{\left(-1{\right)}^{m}{E}_{2m}{\pi }^{2m+1}}{{4}^{m+1}\left(2m\right)!}$

RizerMix

Expert2022-01-27Added 656 answers

My proof works through the following lines: the LHS is: $\frac{1}{\left(2m\right)!}{\int }_{0}^{1}\frac{\right)\mathrm{log}x{\right)}^{2m}}{1+{x}^{2}}dx=\frac{1}{2\cdot \left(2m\right)!}{\int }_{0}^{+\mathrm{\infty }}\frac{\left(\mathrm{log}x{\right)}^{2m}}{1+{x}^{2}}dx$ so we just need to compute: $\frac{{d}^{2m}}{d{k}^{2m}}{\int }_{0}^{+\mathrm{\infty }}\frac{{x}^{k}}{1+{x}^{2}}{|}_{k=0}$ but: ${\int }_{0}^{+\mathrm{\infty }}\frac{{x}^{1/r}}{1+{x}^{2}}dx=r{\int }_{0}^{+\mathrm{\infty }}\frac{{y}^{r}}{1+{y}^{2r}}dy$ $=\frac{\pi /2}{\mathrm{cos}\left(\pi /\left(2r\right)\right)}$ by the residue theorem, so $\sum _{n=0}^{+\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{\left(2n+1{\right)}^{2m+1}}=\frac{{E}_{2m}}{2\cdot \left(2m\right)!}\left(\frac{\pi }{2}{\right)}^{2m+1}$ where ${E}_{2m}$ is just the absolute value of an Euler number, that belongs to $\mathbb{N}$.

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