Find out the answer to "Prove that: ∑n=0∞(−1)n(2n+1)2m+1=(−1)mE2mπ2m+14m+1(2m)!"

spiderifilms6e

Answered question

2022-01-21

Prove that:

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2 m + 1}} = \frac{{(- 1)}^{m} E_{2 m} π^{2 m + 1}}{4^{m + 1} (2 m)!}

nick1337

Expert2022-01-27Added 777 answers

Because it wasnt

star233

Skilled2022-01-27Added 403 answers

The Dirichlet beta function is defined as

β (2 m + 1) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n + 1)^{2 m + 1}}

Then

β (1) = \frac{π}{4}

and

β (2 m + 1) = - \sum_{k = 1}^{m} \frac{(- π^{2} / 4)^{k}}{(2 k)!} β (2 m - 2 k + 1)

(1) If we reindex recursion derived below, we get that the even Euler numbers are defined by

E_{0} = 1

and

E_{2 m} = - \sum_{k = 1}^{m} ((2 m), (2 k)) E_{2 m - 2 k}

(2) then notice that (1) is the same as (2) if we set

β (2 m + 1) = \frac{(- 1)^{m} E_{2 m} π^{2 m + 1}}{4^{m + 1} (2 m)!}

RizerMix

Expert2022-01-27Added 656 answers

My proof works through the following lines: the LHS is:

\frac{1}{(2 m)!} \int_{0}^{1} \frac{) \log x)^{2 m}}{1 + x^{2}} d x = \frac{1}{2 \cdot (2 m)!} \int_{0}^{+ \infty} \frac{(\log x)^{2 m}}{1 + x^{2}} d x

so we just need to compute:

\frac{d^{2 m}}{d k^{2 m}} \int_{0}^{+ \infty} \frac{x^{k}}{1 + x^{2}} |_{k = 0}

but:

\int_{0}^{+ \infty} \frac{x^{1 / r}}{1 + x^{2}} d x = r \int_{0}^{+ \infty} \frac{y^{r}}{1 + y^{2 r}} d y

= \frac{π / 2}{\cos (π / (2 r))}

by the residue theorem, so

\sum_{n = 0}^{+ \infty} \frac{(- 1)^{n}}{(2 n + 1)^{2 m + 1}} = \frac{E_{2 m}}{2 \cdot (2 m)!} (\frac{π}{2})^{2 m + 1}

where

E_{2 m}

is just the absolute value of an Euler number, that belongs to

N

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