What does this series converge to? $\sum_{n=0}^\infty \beta(4n+1)-\beta(4n+3)$

Question

I was investigating patterns in sums of Dirichlet beta function and I found out that if $$a_n=\beta(4n+1)-\beta(4n+3) $$ then the associated series seems to approach a number: $$\sum_{n=0}^\infty a_n=\beta(1)-\beta(3)+\beta(5)-\beta(7)+\beta(9)-\beta(11)+\dots=-0.18698991\dots $$ So my question is: what is the value of this series?

I tried to use the integral representation of $\beta$: $$\beta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}e^{-x}}{1+e^{-2x}}dx $$ but then I couldn't interchange the sum and integral, since this would change the result.

EDIT

As @KStarGamer and @ClaudeLeibovici pointed out, result seems to be $$\sum_{n=0}^\infty a_n\stackrel{(?)}=\frac{\pi}{4}\text{sech}\left(\frac{\pi}{2}\right)-\frac12 $$ Now the question is: how to prove it?

Interesting to look at the partial sum. The coefficients of $\pi^k$ are in $OEIS$ — Claude Leibovici, Feb 08 '24 at 14:20
Is there some reason for which you expect this number to be "something" other than a number? — Marco, Feb 08 '24 at 14:32
Sorry, I don't see any response of @KStarGamer in this thread. — user64494, Feb 08 '24 at 15:13

score 12 · Accepted Answer · answered Feb 08 '24 at 15:20

Recall that the Abel summation $\text{A-}\sum_{n=0}^{\infty} c_n$ is defined by

$$ \text{A-}\sum_{n=0}^{\infty} c_n := \lim_{x \to 1^-} \sum_{n=0}^{\infty} c_n x^n. $$

We collect some properties of Abel summation that is relevant to our computation. (If you are familiar with basic properties of Abel summation, you can skip this part.)

(Abel's Theorem) If $\sum_{n=0}^{\infty} c_n$ is summable in ordinary sense (i.e., convergent as the limit of partial sums), then it is also Abel summable with $$ \text{A-}\sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} c_n. $$
Abel summation is linear. That is, if both $\text{A-}\sum_{n=0}^{\infty} c_n$ and $\text{A-}\sum_{n=0}^{\infty} d_n$ converge and $\alpha$ and $ \beta$ are constants, then $$ \text{A-}\sum_{n=0}^{\infty} (\alpha c_n + \beta d_n) = \alpha \left( \text{A-}\sum_{n=0}^{\infty} c_n \right) + \beta \left( \text{A-}\sum_{n=0}^{\infty} d_n \right) . $$
We have $$ \text{A-}\sum_{n=0}^{\infty} (-1)^n = \frac{1}{2}. $$

Now we return to OP's problem. By noting that $\beta(s) = 1 + \mathcal{O}(3^{-s})$ as $s \to \infty$, we get

\begin{align*} \sum_{n=0}^{\infty} a_n &= \sum_{n=0}^{\infty} (-1)^n [\beta(2n+1) - 1] \\ &= \text{A-}\sum_{n=0}^{\infty} (-1)^n [\beta(2n+1) - 1] \\ &= \text{A-}\sum_{n=0}^{\infty} (-1)^n \beta(2n+1) - \frac{1}{2}. \end{align*}

So it suffices to compute the Abel sum $\text{A-}\sum_{n=0}^{\infty} (-1)^n \beta(2n+1)$. To this end, let $r \in [0, 1)$. Then using the integral representation of $\beta(s)$,

\begin{align*} \sum_{n=0}^{\infty} (-1)^n r^n \beta(2n+1) &= \sum_{n=0}^{\infty} \frac{(-1)^n r^n}{(2n)!} \int_{0}^{\infty} \frac{x^{2n} e^{-x}}{1+e^{-2x}} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{(-1)^n r^n}{(2n)!} x^{2n} \right) \frac{e^{-x}}{1+e^{-2x}} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \cos(\sqrt{r}x) \frac{e^{-x}}{1+e^{-2x}} \, \mathrm{d}x \\ &= \frac{\pi}{4} \operatorname{sech}\left(\frac{\pi \sqrt{r}}{2}\right). \end{align*}

In the second step, we utilized Fubini's Theorem together with the estimate

$$ \int_{0}^{\infty} \sum_{n=0}^{\infty} \left| \frac{(-1)^n r^n}{(2n)!} x^{2n} \right| \frac{e^{-x}}{1+e^{-2x}} \, \mathrm{d}x \leq \int_{0}^{\infty} \cosh(\sqrt{r}x) \frac{e^{-x}}{1+e^{-2x}} \, \mathrm{d}x < \infty. $$

From this, we get

$$ \text{A-}\sum_{n=0}^{\infty} (-1)^n \beta(2n+1) = \frac{\pi}{4} \operatorname{sech}\left(\frac{\pi}{2}\right) $$

and therefore

$$ \sum_{n=0}^{\infty} a_n = \frac{\pi}{4} \operatorname{sech}\left(\frac{\pi}{2}\right) - \frac{1}{2} \approx -0.1869899172\ldots $$

More than impressive ! Thanks for such an answer – Claude Leibovici Feb 08 '24 at 17:19 — Claude Leibovici, Feb 08 '24 at 17:19

score 9 · Answer 2 · answered Feb 08 '24 at 15:23

Recall that $$\beta(s)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}$$ for $\Re(s)>0$. Meaning \begin{align*}S&\!:=\sum_{n=0}^{\infty}\left(\beta(4n+1)-\beta(4n+3)\right)\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{4(-1)^k k (k+1)}{(2k+1)^{4n+3}} = \sum_{n=0}^{\infty}\sum_{k=\color{red}{1}}^{\infty}\frac{4(-1)^k k (k+1)}{(2k+1)^{4n+3}}\\ &=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{4(-1)^k k (k+1)}{(2k+1)^{4n+3}}=\sum_{k=1}^{\infty}\frac{(-1)^k\left(2k+1\right)}{2\left(2k^2+2k+1\right)}.\end{align*} Now recall by the Mittag-Leffler pole expansion for $\sec(z)$, $$\sec(z)=\sum_{k=0}^{\infty}\frac{(-1)^k (2k+1)\pi}{\left(k+\frac{1}{2}\right)^2\pi^2-z^2}.$$ Comparing with our series, we find that $$S=\frac{\pi}{4}\mathrm{sech}\left(\frac{\pi}{2}\right)-\frac{1}{2}.$$

score 4 · Answer 3 · answered Feb 08 '24 at 17:10

PRIMER: PARTIAL FRACTION EXPANSION OF THE SECH FUNCTION

In the Appendix of THIS ANSWER, I showed that

$$\frac{\pi}{2}\sec(\pi a/2)=2\sum_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2-a^2}\tag1$$

Letting $a=i$ in $(1)$, we find that

$$\frac\pi2 \text{sech}(\pi/2)=2\sum_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2+1}\tag2$$

In the following section, we will use $(2)$ to evaluate the series of interest $\sum_{n=0}^\infty \left(\beta(4n+1)-\beta(4n+3)\right)$

Using the series representation of the Beta Function as given by

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

Then, we have

$$\begin{align} \sum_{n=0}^\infty \left(\beta(4n+1)-\beta(4n+3)\right)&=\sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^k \left(\frac{1}{(2k+1)^{4n+1}}-\frac{1}{(2k+1)^{4n+3}}\right)\\\\ &= \sum_{k=1}^\infty (-1)^k \sum_{n=0}^\infty\left(\frac{1}{(2k+1)^{4n+1}}-\frac{1}{(2k+1)^{4n+3}}\right)\\\\ &=\sum_{k=1}^\infty (-1)^k \left(\frac{2k+1}{2(2k^2+2k+1)}\right)\\\\ &=\sum_{k=1}^\infty (-1)^k \frac{2k+1}{ ((2k+1)^2+1)}\\\\ &=\sum_{k=0}^\infty (-1)^k \frac{2k+1}{ ((2k+1)^2+1)}- \frac12\tag3 \end{align}$$

Comparing $(3)$ to $(2)$, we find that

$$\sum_{n=0}^\infty \left(\beta(4n+1)-\beta(4n+3)\right)=\frac\pi4 \text{sech}\left(\frac\pi2\right)-\frac12$$

May I ask why when interchanging the sums the index $k$ starts at $1$ instead of $0$? — Zima, Feb 08 '24 at 19:03
@Zima Of course. Note the inner sum over $k$. When $k=0$, the summand is equal to $0$. So, we can exclude this. And in fact, we need to do so in order to apply the sum of a convergent geometric series. Note that we need $2k+1>1\implies k>0$. So we can exclude $k=0$ and we must exclude $k=0$. — Mark Viola, Feb 08 '24 at 19:51

Random Variable · Answer 4 · 2024-02-13T18:42:28.847

The series can be written as

$$ \begin{align} \sum_{n=0}^{\infty} \left(\beta(4n+1)-\beta(4n+3) \right) &= \sum_{n=0}^{\infty} \big(\left(\beta(4n+1)-1\right) - \left(\beta(4n+3)-1\right) \big) \\ &= \sum_{n=0}^{\infty} (-1)^{n} \left(\beta(2n+1)-1 \right). \end{align}$$

(The series is not $\sum_{n=0}^{\infty} (-1)^{n} \beta(2n+1) $, which is not a convergent series but exists in the Abel sense as Sangchul Lee showed. Wolfram Alpha, however, is strangely insistent that it does converge. )

For $s>0$, an integral representation for $\beta(s)-1$ is $$ \begin{align} \beta(s) -1 &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-2x}} \, \mathrm dx - \frac{1}{\Gamma(s)} \int_{0}^{\infty} x^{s-1}e^{-x} \, \mathrm dx \\ &= -\frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}e^{-3x}}{1+e^{-2x}} \, \mathrm dx. \end{align}$$

Therefore, we have $$ \begin{align} \sum_{n=0}^{\infty} (-1)^{n} \left(\beta(2n+1)-1 \right) &= -\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{0}^{\infty} \frac{x^{2n} e^{-3x}}{1+e^{-2x}} \, \mathrm dx \\ &= -\int_{0}^{\infty} \frac{e^{-3x}}{1+e^{-2x}} \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n}}{(2n)!} \, \mathrm dx \\ &= -\int_{0}^{\infty} \frac{e^{-3x}\cos(x)}{1+e^{-2x}} \, \mathrm dx\\ &= - \int_{0}^{\infty} \left(e^{-x}- \frac{1}{2}\frac{1}{\cosh(x)} \right)\cos(x) \, \mathrm dx \\ &=- \int_{0}^{\infty} e^{-x} \cos(x) \, \mathrm dx + \frac{1}{2} \int_{0}^{\infty} \frac{\cos (x)}{\cosh (x)} \, \mathrm dx \\ &\overset{\spadesuit}{=} - \frac{1}{2} + \frac{\pi}{4} \, \operatorname{sech} \left(\frac{\pi}{2} \right). \end{align}$$

Switching the order of integration and summation was justified by Fubini's theorem since $$\int_{0}^{\infty} \sum_{n=0}^{\infty} \left|-\frac{e^{-3x}}{1+e^{-2x}} \frac{(-1)^{n}x^{2n}}{(2n)!} \right| \, \mathrm dx = \int_{0}^{\infty} \frac{e^{-3x} \cosh(x)}{1+e^{-2x}} < \infty . $$

$\spadesuit $ Solve $\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx$ without complex integration.

Fourier transform of 1/cosh

We also have $$\sum_{n=0}^{\infty} \left(\beta(2n+1)-1 \right) = -\int_{0}^{\infty} \frac{e^{-3x} \cosh(x)}{1+e^{-2x}} \, \mathrm dx = -\frac{1}{2} \int_{0}^{\infty} e^{-2x} \, \mathrm dx = -\frac{1}{4}, $$

$$\sum_{n=1}^{\infty} \left(\beta(n)-1 \right) = - \int_{0}^{\infty} \frac{e^{-2x}}{1+e^{-2x}} \, \mathrm dx = - \frac{1}{2} \int_{0}^{1} \frac{\mathrm du}{u} = -\frac{\ln(2)}{2},$$

and

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\left(\beta(n)-1 \right) = -\int_{0}^{\infty} \frac{e^{-4x}}{1+e^{-2x}} \, \mathrm dx = \frac{1}{2} \int_{0}^{1} \left( \frac{1}{1+u}-1 \right) \mathrm du = \frac{1}{2} \left(\ln(2)-1 \right).$$

For some reason Wolfram Alpha won't return approximate values for any of these series unless you remove the first term, which is just $\beta(1) -1 = \frac{\pi}{4}-1$.

xpaul · Answer 5 · 2024-02-08T16:46:46.133

Note $$\begin{eqnarray} S&=&\sum_{n=0}^\infty(\beta(4n+1)-\beta(4n+3))=\sum_{k=0}^\infty(-1)^k\beta(2k+1) \\ &=&\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{(-1)^{k+n}}{(2n+1)^{2k+1}}=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{(-1)^{k+n}}{(2n+1)^{2k+1}}\\ &=&\sum_{n=0}^\infty(-1)^n\sum_{k=0}^\infty\frac{(-1)^{k}}{(2n+1)^{2k+1}}=\frac14\sum_{n=0}^\infty(-1)^n\frac{2n+1}{n^2+n+\frac12}\\ &=&\frac14\sum_{n=0}^\infty(-1)^n\bigg[\frac{1}{n+\frac12(1-i)}+\frac{1}{n+\frac12(1+i)}\bigg]\\ &=&\frac14\sum_{n=0}^\infty\bigg[\bigg(\frac{1}{2n+\frac12(1-i)}-\frac{1}{2n+1+\frac12(1-i)}\bigg)+\bigg(\frac{1}{2n+\frac12(1+i)}-\frac{1}{2n+1+\frac12(1+i)}\bigg)\bigg]\\ &=&\frac18\sum_{n=0}^\infty\bigg[\bigg(\frac{1}{n+\frac14-\frac i4}-\frac{1}{n+\frac34-\frac i4}\bigg)+\bigg(\frac{1}{n+\frac14+\frac i4}-\frac{1}{n+\frac34+\frac i4}\bigg)\bigg]\\ &=&\frac18\sum_{n=0}^\infty\bigg[\bigg(\frac{1}{n+\frac14-\frac i4}-\frac{1}{n+\frac34+\frac i4}\bigg)+\bigg(\frac{1}{n+\frac14+\frac i4}-\frac{1}{n+\frac34-\frac i4}\bigg)\bigg]\\ \end{eqnarray}$$ Using $$ \sum_{n=0}^\infty\bigg(\frac1{n\alpha+\beta}-\frac1{n\alpha+(\alpha-\beta)}\bigg)=\frac{\pi}{\alpha}\cot(\frac{\pi\beta}{\alpha}) $$ one has $$ S=\frac18\bigg(i\pi\coth(\frac{1+i}{4})+\pi\cot(\frac{1+i}{4})\bigg)=2\pi\text{sech}(\frac{\pi}2). $$

user64494 · Answer 6 · 2024-02-08T14:57:29.540

2

Mathematica 14.0 gives the closed-form answer by

ComplexExpand[Sum[(Integrate[1/Gamma[s]*x^(s - 1)*Exp[-x]/(1 + Exp[-2 x]) /. s -> 4 n + 1, {x,
   0, Infinity}]) - Integrate[
1/Gamma[s]*x^(s - 1)*Exp[-x]/(1 + Exp[-2 x]) /. s -> 4 n + 3, {x, 
 0, Infinity}], {n, 0, Infinity}] // FullSimplify]

$$\frac{\pi \cosh \left(\frac{\pi }{2}\right)}{2 (1+\cosh (\pi ))} $$

edited Feb 08 '24 at 14:57

answered Feb 08 '24 at 14:39

user64494

5,811

I overlooked that summation begins from $0$. See the edited answer. – user64494 Feb 08 '24 at 14:58
$-0.18698991\dots$ in the question is not grounded. – user64494 Feb 08 '24 at 15:02
@KStarGamer. Great ! Please, put an answer with the Mma syntax – Claude Leibovici Feb 08 '24 at 15:04
This is also $\frac{\pi}{4} \text{sech}\left(\frac{\pi }{2}\right)-\frac{1}{2}$ which is shorter !! – Claude Leibovici Feb 08 '24 at 15:08
Mathematica has a problem with $n=0$. – user64494 Feb 08 '24 at 15:33

What does this series converge to? $\sum_{n=0}^\infty \beta(4n+1)-\beta(4n+3)$

6 Answers6