I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?

  • $\begingroup$ You might want to try using the $E_{2m}$ have $\sec x$ as its EGF. $\endgroup$ – Pedro Tamaroff May 7 '14 at 4:19
  • $\begingroup$ @PedroTamaroff: the Euler numbers used above have $\mathrm{sech}(x)$ as their EGF, but that is what I used in my derivation of the recursion in my answer. $\endgroup$ – robjohn May 10 '14 at 0:37

Because it wasn't explicitly stated, I probably should mention that $E_{2m}$ are the Euler numbers.

It can be shown using contour integration.

Let $ \displaystyle f(z) = \frac{\pi \csc \pi z}{(2z+1)^{2m+1}}$.

Then integrating around a square with vertices at $\pm (N+\frac{1}{2}) \pm i(N+\frac{1}{2})$ (where $N$ is a positive integer greater than $1$) and letting $N$ go to infinity,

$$ \sum_{n=-\infty}^{\infty} \text{Res}[f(z),n] + \text{Res} \left[f(z),- \frac{1}{2} \right] $$

$$ = 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} + \text{Res}\left[f(z),-\frac{1}{2} \right] = 0 .$$

But notice that

$$\text{Res} \Big[ f(z),- \frac{1}{2} \Big] = \text{Res} \Big[f (z - \frac{1}{2}),0 \Big] = \text{Res} \Big[\frac{-\pi \sec \pi z}{(2z)^{2m+1}},0 \Big] .$$

And using the Taylor expansion of sec(z) at the origin, that is, $$ \sec z = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k}$$

we have

$$\frac{-\pi \sec \pi z}{(2z)^{2m+1}} = \frac{-\pi}{2^{2m+1}} \sum_{k=0}^{\infty} \frac{(-1)^{k} E_{2k}}{(2k)!} \pi^{2k} z^{2k-2m-1} ,$$

which implies

$$\text{Res} \Big[\frac{-\pi \sec \pi z}{(2z)^{2m+1}},0 \Big] = \frac{-\pi (-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m} .$$


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

and the result follows.


Another approach is to use the partial fractions expansion

$$ \sec(z) = \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1) \pi}{\left(\frac{2n+1}{2} \right)^{2} \pi^{2}-z^{2}} $$


$$ \begin{align}\sec(z) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1)\pi}{\left(\frac{2n+1}{2} \right)^{2}\pi^{2}} \frac{1}{1-\left( \frac{2z}{(2n+1)\pi}\right)^{2}} \\ &=\frac{4}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=0}^{\infty} \frac{(2z)^{2k}}{\left((2n+1) \pi\right)^{2k}} \ \ (|z| < \frac{\pi}{2}) \\ &= \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} \\ &= \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \beta(2k+1) \ .\end{align} $$

But we already know that

$$ \sec(z) = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k} .$$


$$ \frac{4}{\pi} \frac{2^{2k}}{\pi^{2k}} \beta(2k+1) = \frac{(-1)^{k} E_{2k}}{(2k)!}$$

which implies

$$ \begin{align}\beta(2k+1) &= \frac{(-1)^{k} E_{2k} }{2^{2k+2}(2k)!} \pi^{2k+1} \\ &= \frac{(-1)^{k} E_{2k} }{4^{k+1}(2k)!} \pi^{2k+1} . \end{align}$$

  • $\begingroup$ @Peter I added a second proof to my answer. $\endgroup$ – Random Variable May 7 '14 at 4:19
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    $\begingroup$ $\sum_{n=0}^\infty\sum_{k=0}^\infty|(-1)^n\frac{z^{2k}}{{(2n+1)^{2k+1}}}|$ is divergent, How did you switch the order of summation here? $\endgroup$ – user91500 May 9 '14 at 14:11
  • $\begingroup$ @user91500 The justification might be nontrivial. That's why I didn't post that second evaluation until Peter asked about a different approach. Sufficient, but not necessary, conditions is if $$\sum_{n=0}^{\infty} \sum_{k=0}^{\infty}|\frac{2^{2k} z^{2k}}{\pi^{2k}} \frac{(-1)^{n}}{(2n+1)^{2k+1}}| $$ or $$\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} |\frac{2^{2k} z^{2k}}{\pi^{2k}} \frac{(-1)^{n}}{(2n+1)^{2k+1}}|$$converges. The second iterated sum does converge if $k$ starts from $k=1$. That might show that the sum is well-behaved enough to switch the order of summation. Let me ask around. $\endgroup$ – Random Variable May 9 '14 at 15:45
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    $\begingroup$ @user91500: you can pull out the $k=0$ term and handle it separately. In both sums that is $$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$ Then the double sums for $n\ge0$ and $k\ge1$ converge absolutely for $|z|\lt\frac\pi2$. $\endgroup$ – robjohn May 9 '14 at 16:38
  • $\begingroup$ @robjohn Thanks. $\endgroup$ – Random Variable May 9 '14 at 16:44

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(6)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$


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


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