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-


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 $B$.


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

A list of a few terms would be







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

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


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


Beyond this observation I am clueless.


1 Answer 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 $\approx 1.9$.

  • $\begingroup$ @RounakSarkar Haha, I wasn't expecting it to work out so nicely given how complicated the problem looks! :D $\endgroup$ Oct 11, 2021 at 17:00
  • $\begingroup$ @metamorphy Oops, thanks for fixing that :) $\endgroup$ Oct 12, 2021 at 13:15

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