First of all, I would like to thank you for the consultation and the opportunity to ask for help on very serious issues.
I'll go straight to the question itself.
Task:
It is necessary to calculate the sum of the following row:
$$
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b}
$$
To begin with, I decided to calculate the value of this sum directly using the program (Maple or Wolfram Mathematica):
S:=Sum(Sum( a^2*b^2/sinh(Pi*(a+b))*(-1)^(a+b), a=1..infinity), b=1..infinity);
S=identify(%);
out[1]: S = (1)/(120*\pi) = 0.00265
Next, I try to prove that the given series (sum) really tends to take on the value calculated using the program.
Proof:
Now, let me present your attempt at proof. And I'm experiencing some specific difficulties. Those. I am trying to calculate the given amount, first the inner, then the outer, as follows:
\begin{align*}
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b} &= \\
& = \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi(a+b))}(-1)^{b} \\ \ &= \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \left( \frac{\pi^2}{6} + \sum_{b=2}^{\infty} \frac{b^2} {\sinh(\pi b + \pi a)}(-1)^{b} \right) \ \\
&= \frac{\pi^2}{6} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=2}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b} \\\
&= \frac{\pi^2}{6} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \left( \frac{\pi^2}{6} – 1 + \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b} \right) \ \\
&= \frac{\pi^2}{6} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \frac{\pi^2}{6} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} – \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b} \\
&= \frac{\pi^2}{3} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b}
\end{align*}
After, let consider two sums separately. Consider the first sum
$$
\frac{\pi^2}{3} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a}
$$
The first sum is a series of the form $\frac{\pi^2}{3}\sum_{a=1}^{\infty} \frac{a^2}{(-1)^a}$ where the terms are alternating and the denominator is $(-1)^a$. This means that the series is conditionally convergent, meaning that the sum depends on the order in which the terms are added.
The second sum is an infinite geometric series, the general formula of this series is $\frac{a}{1-r}$ where a is the first term and r is the common ratio. In this case, the first term is 1 and the common ratio is 1, so the second sum is equal to $\frac{1}{1-1}$ which is divergent.
Thus, the overall sum is divergent and it doesn't have a finite sum.
Consider the second sum
$$
\sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b}
$$
This sum is a combination of two infinite series. The first series is a conditionally convergent series and the second series is a divergent series.
It's worth noting that the sum of two divergent series doesn't have to be divergent, but in this case, both series are divergent and the sum of them is also divergent.
Finally, this sum
$$
\frac{\pi^2}{3} \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} + \sum_{a=1}^{\infty} \frac{a^2}{(-1)^a} \sum_{b=1}^{\infty} \frac{b^2}{\sinh(\pi b + \pi a)}(-1)^{b}
$$
not converge, and so, this sum not have a limit and I don't understand how get the values $\frac{1}{120 \cdot \pi}$.
Taking into account the above, I have a question – is it possible to prove that this sum has a limit and converges, or does this multiple series diverge, and if it diverges, then how can this be justified more correctly?
Edit 1:
I think, that
$$
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b}
$$
is equal to
$$
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} (-1)^{a+b} \ a^2b^2 \ 2e^{(a+b)(-1-2k)\pi}
$$
where
$$
\frac{1}{\sinh(\pi(a+b))} = \sum_{k=0}^{\infty} 2e^{(a+b)(-1-2k)\pi}
$$
Using Mathematica I obtain same solution:
$$
\frac{2e^{2(1-2k)\pi}(-1+e^{\pi+2\pi k})^2}{(1+e^{\pi+2\pi k})^6}
$$
Though, I don't understand, how it works.
So, after using Mathematica I evaluate sum:
$$
\sum_{k=0}^{\infty}\frac{2e^{2(1-2k)\pi}(-1+e^{\pi+2\pi k})^2}{(1+e^{\pi+2\pi k})^6}
$$
and get
$$
\frac{1}{30\pi} +\frac{\psi_q(1,0.5-0.5i,e^{2\pi})}{60\pi^2} + \frac{\psi_q(5,0.5-0.5i,e^{2\pi})}{900\pi^6} (*)
$$
where
$$
\psi_q(n,z,q)=-ln(1-q)+ln(q) \sum_{n}^{\infty}\frac{(q^{n+z})}{(1-q^{n+z})}.
$$
And if simplify (*) I get $\frac{1}{120\pi}$. I am frustrated and very discouraged, because I don’t understand why such a sum, and where does the digamma function come from in the expansion.
Have any idea for solution this problem?
Best Answer
Pairing up all $a,\,b$ that satisfy $a+b=n$ gives
\begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b} &= \sum_{n=2}^\infty\frac{(-1)^n}{\sinh(\pi n)}\sum_{k=1}^{n-1}k^2(n-k)^2 \\ &= \frac{1}{30}\sum_{n=2}^\infty\frac{(n^5-n)(-1)^n}{\sinh(\pi n)} \\ &= \frac{1}{60}\sum_{n\in\mathbb{Z}\setminus\left\{0\right\}}\frac{(n^5-n)(-1)^n}{\sinh(\pi n)}. \end{align*}
We now make use of the residue theorem to evaluate the above bilateral infinite series. Let $C_L$ be a positively oriented square contour of side length $L$ centered at $z=0$ with vertices at $z=(\pm1\pm i)L/2$. As $L\to\infty$, one may bound the integrals along the edges to show that
\begin{align} \lim_{L\to\infty}\frac{1}{2\pi i}\oint_{C_L}\frac{(z^5-z)}{60\sinh(\pi z)}\pi\csc(\pi z)\,\mathrm dz=0. \end{align}
(Details relevant to the above method may be seen here.) The integrand has poles at $\mathbb{Z}$ and $i\mathbb{Z}$, set minus $\pm1$ and $\pm i$, applying the residue theorem tells us
\begin{align} 0 &= \underset{z=0}{\text{Res}}\left[\frac{(z^5-z)}{60\sinh(\pi z)}\pi\csc(\pi z)\right]+\sum_{n\in\mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\underset{z=n}{\text{Res}}\left[\frac{(z^5-z)}{60\sinh(\pi z)}\pi\csc(\pi z)\right] \\ &\qquad+\sum_{n\in \mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\underset{z=in}{\text{Res}}\left[\frac{(z^5-z)}{60\sinh(\pi z)}\pi\csc(\pi z)\right] \\ &=-\frac{1}{60\pi}+\frac{1}{60}\sum_{n\in\mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\frac{(n^5-n)(-1)^n}{\sinh(\pi n)}+i\sum_{n\in \mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\underset{z=n}{\text{Res}}\left[\frac{((iz)^5-iz)}{60\sinh(\pi iz)}\pi\csc(\pi iz)\right] \\ &= -\frac{1}{60\pi}+\frac{1}{60}\sum_{n\in\mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\frac{(n^5-n)(-1)^n}{\sinh(\pi n)}+\sum_{n\in\mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\underset{z=n}{\text{Res}}\left[\frac{(z^5-z)}{60\sinh(\pi z)}\pi\csc(\pi z)\right] \\ &= -\frac{1}{60\pi}+\frac{1}{30}\sum_{n\in\mathbb{Z}\setminus\left\{0,\,\pm1\right\}}\frac{(n^5-n)(-1)^n}{\sinh(\pi n)}. \end{align}
This implies that
$$\frac{1}{60}\sum_{n\in\mathbb{Z}\setminus\left\{0\right\}}\frac{(n^5-n)(-1)^n}{\sinh(\pi n)}=\frac{1}{120\pi}$$
or put differently,
$$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b}=\frac{1}{120\pi}.$$