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
    May 7, 2014 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, 2014 at 0:37

5 Answers 5


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

Let's integrate the function $$ f(z) = \frac{\pi \csc (\pi z)}{(2z+1)^{2m+1}}, \quad m \in \mathbb{N}_{\ge 0},$$

around a square contour with vertices at $\pm (N+\frac{1}{2}) \pm i(N+\frac{1}{2})$ , where $N$ is a positive integer.

Letting $N$ go to infinity throught the positive integers, the integral vanishes (see here), and we end up with

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

Notice that

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

Using the Maclaurin series of sec(z), that is, $$ \sec (z) = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!}z^{2n} \, , \quad |x| < \frac{\pi}{2},$$

we have

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

from which we can conclude that

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


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

and the result follows.


Another approach is to use the partial fractions expansion

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

For $|z| < \frac{\pi}{2}$, we have

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

But we know that

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


$$ \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k},$$

and the result follows by comparing the $k$th coefficient of both series.

  • $\begingroup$ @Peter I added a second proof to my answer. $\endgroup$ May 7, 2014 at 4:19
  • 1
    $\begingroup$ $\sum_{n=0}^\infty\sum_{k=0}^\infty|(-1)^n\frac{z^{2k}}{{(2n+1)^{2k+1}}}|$ is divergent, How did you switch the order of summation here? $\endgroup$
    – user91500
    May 9, 2014 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. $\endgroup$ May 9, 2014 at 15:45
  • 1
    $\begingroup$ @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, 2014 at 16:38
  • $\begingroup$ @robjohn Thanks. $\endgroup$ May 9, 2014 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 $\mathbb{N}$.


\begin{gather*} \beta(2m+1)=\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x=\frac{1}{(2m)!}\left(\int_0^\infty-\int_1^\infty\right)\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\ =\frac{1}{(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x-\frac{1}{(2m)!}\underbrace{\int_1^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x}_{x\to 1/x}\\ =\frac{1}{(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x-\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\ \left\{\text{add $\beta(2m+1):=\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x$ to both sides then divide by $2$}\right\}\\ =\frac{1}{2(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\ =\frac{\pi}{4^{m+1}(2m)!}\lim_{s\to \frac12}\frac{d^{2m}}{ds^{2m}}\csc(\pi s). \end{gather*}

We showed here

$$\lim_{s\to \frac12}\frac{d^{2m}}{ds^{2m}}\csc(s\pi)=|E_{2m}|\pi^{2m}.$$ Using this result, it follows that $$\beta(2m+1)=\frac{|E_{2m}|\pi^{2m+1}}{4^{m+1}(2m)!}.$$


Polygamma Function

The polygamma function of order $(m-1)$ is defined as the $m$-th derivative of $(\ln \Gamma(x))$:

$$ \psi^{(m-1)}(x) = \frac{d^m}{dx^m} \ln \Gamma(x) $$

Reflection Relation

The reflection relation between polygamma functions is given by

$$ (-1)^m \psi^{(m)}(1-z) - \psi^{(m)}(z) = \pi \frac{\mathrm{d}^m}{\mathrm{d} z^m} \cot(\pi z). $$

It can be easily derived by differentiating the reflection relation of the gamma function $m$ times:

$$ \frac{d^m}{dz^m} \ln\left(\Gamma(z) \Gamma(1-z)\right) = \frac{d^m}{dz^m} \ln\left(\frac{\pi}{\sin(\pi z)}\right) $$

Dirichlet Beta Function

The Dirichlet beta function is defined as:

$$ \beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} $$

If we separate terms like this:

$$ \beta(s) = \sum_{n=0}^\infty \frac{1}{(4n + 1)^s} - \sum_{n=0}^\infty \frac{1}{(4n + 3)^s} $$

Then, it can be expressed in terms of the Hurwitz zeta function and polygamma function as well:

$$ \beta(s) = 4^{-s} \left( \zeta\left(s,\frac{1}{4}\right)-\zeta\left(s,\frac{3}{4}\right) \right) $$

Polygamma series expansion

For the polygamma function, let us recall(derive) the series expansion:

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \ln \Gamma(x) $$ diffrentiating

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \sum_{k=1}^{x} \frac{1}{k} $$ adding and subtracting $ \sum_{k=x+1}^{\infty} \frac{1}{k} $ of eqn

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{x} \frac{1}{k}+ \sum_{k=x+1}^{\infty} \frac{1}{k} - \sum_{k=x+1}^{\infty} \frac{1}{k}\right) $$ now $$ \psi^{(m)}(x) =\frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{x+k}\right) $$

now we can see that first term inside brackets will be treated as constant because its converging and we can easily prove it by integral test or also approximate for more info see Harmonic Series and by differentiating it m times we get below expression $$ \psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}} $$

Now, considering specific values for $z = 1/4$ and $z = 3/4$:

$$ \psi^{(m)}\left(\frac{1}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{1}{4}+k\right)^{m+1}} $$

$$ \psi^{(m)}\left(\frac{3}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{3}{4}+k\right)^{m+1}} $$

Then, subtracting these two equations gives us the relation:

$$ \beta(s) = \frac{1}{2^s} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{s}} = \frac1{(-4)^s(s-1)!}\left[\psi^{(s-1)}\left(\frac{1}{4}\right)-\psi^{(s-1)}\left(\frac{3}{4}\right)\right] $$

If (s) is odd, then:

$$ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac1{(-4)^{2m+1}(2m)!}\left[\psi^{(2m)}\left(\frac{1}{4}\right)-\psi^{(2m)}\left(\frac{3}{4}\right)\right] $$

By using the reflection formula of the polygamma function, we obtain the value of:

$$ (-1)^m \psi^{(m)}\left(1-\frac{1}{4}\right) - \psi^{(m)}\left(\frac{1}{4}\right) = \left. \frac{\pi \, \mathrm{d}^{(m)}}{\mathrm{d} z^{(m)}} \cot(\pi z) \right|_{z=\frac{1}{4}} $$

Which gives us the formula:

$$ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac{1}{(4)^{2m+1}(2m)!}\Bigg|_{z=\frac{1}{4}} \frac{\pi \, \mathrm{d}^{(2s)}}{\mathrm{d} z^{(2s)}} \cot(\pi z) $$


it was solution by me posted here


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