Real Analysis – Convergence of Series $\sum_{n=0}^\infty \beta(4n+1)-\beta(4n+3)$

calculusintegrationlimitsreal-analysissequences-and-series

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?

Best Answer

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.)

  1. (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 $$ \text{A-}\sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} c_n. $$

  2. 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 $$ \text{A-}\sum_{n=0}^{\infty} (\alpha c_n + \beta d_n) = \alpha \left( \text{A-}\sum_{n=0}^{\infty} c_n \right) + \beta \left( \text{A-}\sum_{n=0}^{\infty} d_n \right) . $$

  3. 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

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

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