Closed-form for $B=\lim_{n\to\infty}\sum_{a_1=1}^{\infty}\frac{1}{a_1^2}\sum_{a_2=1}^{a_1}\frac{1}{a_2^2}\cdots\sum_{a_n=1}^{a_{n-1}}\frac{1}{a_n^2}$?

Introduction

This problem came to my mind few years ago when I first learned about limits and infinite sums. I saw sums, double sums, triple sums etc, but never an infinite sum basically an infinite chain of summation symbols which will converge to a real value. So I constructed the expression below,

My Question

Is there a closed form of the following number-

$\textstyle\displaystyle{B=\lim_{n\rightarrow\infty}\sum_{a_1=1}^{\infty}\frac{1}{a_1^2}\sum_{a_2=1}^{a_1}\frac{1}{a_2^2}\cdots\sum_{a_n=1}^{a_{n-1}}\frac{1}{a_n^2}}$

Or some more easier representations not involving this infinite chain of sigma symbols with only finite amount of indices? Does it even converge?

Details And Observation

Let's first define a double sequence for which will converge to $B$.

$\textstyle\displaystyle{\Sigma_{m,n}=\sum_{a_1=1}^{n}\frac{1}{a_1^2}\sum_{a_2=1}^{a_1}\frac{1}{a_2^2}\cdots\sum_{a_m=1}^{a_{m-1}}\frac{1}{a_m^2}}$

So, $B=\Sigma_{\infty,\infty}$

A list of a few terms would be

$\Sigma_{1,\infty}$$\textstyle\displaystyle{=\frac{\pi^2}{6}}$

$\Sigma_{2,\infty}$$\textstyle\displaystyle{=\frac{7\pi^4}{360}}$

$\Sigma_{3,100}$$\textstyle\displaystyle{=1.95233691...}$

$\Sigma_{4,36}$$\textstyle\displaystyle{=1.93916539...}$

$\Sigma_{5,28}$$\textstyle\displaystyle{=1.92936096...}$

$\vdots$

I can't get more values for this, Wolfram Alpha doesn't work.

Notice that $\Sigma_{m,n}$ satisfies a recurrence relation-

$\textstyle\displaystyle{\Sigma_{m,n}=\sum_{k=1}^{n}\frac{\Sigma_{m-1,k}}{k^2}}$

Where, $\Sigma_{0,n}=1$

This recurrence relation is very similar to Hyperharmonic numbers, it just has a factor of $k^{-2}$ within the sum and the initial value is different. The hyperharmonic numbers satisfies

$\textstyle\displaystyle{H_n^{(r)}=\sum_{k=1}^{n}H_k^{(r-1)}}$

Beyond this observation I am clueless.

Solution 1:

Let us define

$$\Sigma_{\infty,n} = \lim_{m\to\infty} \Sigma_{m,n}$$

It can be proven using simple induction that $\Sigma_{m,1}=1$ for all $m\in\mathbb N$, hence we have that $\Sigma_{ \infty,1}=1$. Now, from your recurrence relation, taking the limit of both sides as $m\to\infty$, we have

$$\Sigma_{\infty,n}=\sum_{k=1}^n \frac{\Sigma_{\infty,k}}{k^2}=\Sigma_{\infty,n-1}+\frac{\Sigma_{\infty, n}}{n^2}$$

which may be solved for $\Sigma_{\infty,n}$, yielding

$$\Sigma_{\infty, n}=\frac{\Sigma_{\infty, n-1}}{1-\frac{1}{n^2}}$$

Now, since we already have that $\Sigma_{\infty, 1}$, we obtain the following formula by induction:

$$\Sigma_{\infty, n} = \prod_{k=2}^n \Big(1-\frac{1}{k^2}\Big)^{-1}$$

which gives us the desired result:

$$\color{green}{\Sigma_{\infty,\infty} = \prod_{k=2}^\infty \Big(1-\frac{1}{k^2}\Big)^{-1} = 2}$$

which is supported by your numerical evidence, showing values $\approx 1.9$.