Undergrad Infinite Series of Infinite Series

[Drakkith]

TL;DR

How to solve an infinite series where each term is itself an infinite series?

I had a random thought about infinite series the other day while watching a math video. Let's say we have an infinite series where each term in the series is itself another infinite series. How would one go about finding the sum?

For example, let's say we have the series $a_1+a_2+a_3...$ where $a_n = (\frac{1}{2})^n+(\frac{1}{4})^n+(\frac{1}{8})^n...$
I assume the series converges since each term is the reciprocal powers of two series raised to a power (without the leading 1/1 term, which would make it diverge), which should fall off quite quickly. However it's been a while since I did any math work involving series so I'm a bit unsure. Thoughts? Is solving something like this fundamentally any different from solving a 'plain' series?

 

[jedishrfu]

Sometimes one can reorganize the series by grabbing the first term of each sub series and group them as the first term of a new series.

 

[renormalize]

For example, let's say we have the series $a_1+a_2+a_3...$ where $a_n = (\frac{1}{2})^n+(\frac{1}{4})^n+(\frac{1}{8})^n...$

This is a double sum (akin to a double integral) that we may call $S$. The inner sum is trivial and Mathematica can compute the remaining sum:$$S\equiv\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}2^{-mn}=\sum_{m=1}^{\infty}\left(\frac{1}{2^{m}-1}\right)=1-\frac{\psi_{1/2}^{0}\left(1\right)}{\ln2}$$where $\psi_{q}^{n}\left(z\right)$ is the q-Polygamma Function.

 

[Drakkith]

Excellent! I would have never thought about the series being like a double integral! Thanks so much!

 

[pbuk]

Mathematica can compute the remaining sum...

But don't leave us in suspense: the result is the Erdős–Borwein constant (approximately 1.6) and, remarkably, is also equal to $$ \sum_{n=1}^{\infty} \frac{\sigma_0(n)}{2^n} $$ where $\sigma_0(n)$ is the number of divisors of $n$.

 

[fresh_42]

TL;DR Summary: How to solve an infinite series where each term is itself an infinite series?

Is solving something like this fundamentally any different from solving a 'plain' series?

You only have to be careful in case you re-order the summations, e.g. swapping the sums or replacing one sum with an integral and swapping sum and integration. In such cases, a phenomenon called "mass vanishing at infinity" can occur. For example, consider the telescope sum
$$
\sum_{k=0}^n\underbrace{
\left(\chi_{[k,k+1]}-\chi_{[k+1,k+2]}\right)
}_{=f_k}=\sum_{k=0}^n \left(\chi_{[0,1]}-\chi_{[n+1,n+2]}\right)\stackrel{n\to \infty }{\longrightarrow }\chi_{[0,1]}
$$
then we have
$$
0=\sum_{k=0}^\infty \left(\int_\mathbb{R}f_k(x)\,dx\right)\neq \int_\mathbb{R} \left(\sum_{k=0}^\infty f_k(x)\right)\,dx=1
$$
This isn't a problem in your example. I just wanted to mention that it can be tricky if two infinite sums are involved.

 

[fresh_42]

But don't leave us in suspense: the result is the Erdős–Borwein constant (approximately 1.6) and, remarkably, is also equal to $$ \sum_{n=1}^{\infty} \frac{\sigma_0(n)}{2^n} $$ where $\sigma_0(n)$ is the number of divisors of $n$.

And here is the link to the Wikipedia article:
https://en.wikipedia.org/wiki/Erdős–Borwein_constant
I found it interesting that it actually occurs somewhere:

The Erdős–Borwein constant comes up in the average case analysis of the heapsort algorithm, where it controls the constant factor in the running time for converting an unsorted array of items into a heap.

 

[Drakkith]

Well, I'm glad I picked an interesting example!