Infinite series that surprisingly converge?

Question

I couldn't find any substantial list of 'strange infinite convergent series' so I wanted to ask the MSE community for some. By strange, I mean infinite series/limits that converge when you would not expect them to and/or converge to something you would not expect.

My favorite converges to Khinchin's (sometimes Khintchine's) constant, $K$. For almost all $x \in \mathbb{R}$ (those for which this does not hold making up a measure zero subset) with infinite c.f. representation: $$x = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac1{\ddots}}}$$ We have: $$\lim_{n \to \infty} =\root n \of{\prod_{i=1}^na_i} = \lim_{n \to \infty}\root n \of {a_1a_2\dots a_n} = K$$ Which is...wow! That it converges independent of $x$ really gets me.

Answer 1

A pretty commonly mentioned one is the Kempner series, which is the Harmonic series but "throwing out" (omitting) the numbers with a 9 in their decimal expansion. And 9 is not special; you can generalize to any finite sequence of digits, and the series will converge. MathWorld has approximate values for the single-digit possibilities.

Answer 2

I still like the fact that $$ \sum_{n=N}^\infty \frac{1}{n\ln n \cdot \ln \ln n \cdot \ln \ln \ln n \cdot \ln \ln \ln \ln n} $$ diverges, but $$ \sum_{n=N}^\infty \frac{1}{n\ln n \cdot \ln \ln n \cdot \ln \ln \ln n \cdot (\ln \ln \ln \ln n)^{1.01}} $$ converges (where $N$ is a large enough constant for the denominator to be defined).

Answer 3

I would like to nominate an infinite product:

$\prod_{n=2}^{\infty}\dfrac{n^3-1}{n^3+1}=\dfrac{2}{3}$

Proof: Factor thusly:

$n^3-1=(n-1)(n^2+n+1)=((n-2)+1)(n^2+n+1)$

$n^3+1=(n+1)(n^2-n+1)=(n+1)((n-1)^2+(n-1)+1)$

and the product then telescopes.

Answer 4

Let $x_n$ be the nth positive solution of $\csc(x)=x$, i.e. $x_1\approx 1.1141$, $x_2\approx 2.7726$, etc. Then,

$$\sum_{n=1}^{\infty}\frac{1}{x_n^2}=1$$

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Edit: Even more surprisingly, if we define $s(k)=\sum x_n^{-k}$, then we have the generating function

\begin{align*} \sum_{k=1}^{\infty}s(2k)x^{2k} &=\frac{x}{2}\left(\frac{1+x\cot(x)}{\csc(x)-x}\right) \\ &=x^2+\frac{2x^4}{3}+\frac{21x^6}{40}+\frac{59x^8}{140}+\frac{24625x^{10}}{72576}+\cdots \end{align*}

Unfortunately it seems that, as with the Riemann zeta function, the values of $s$ at odd integers are out of reach.

Answer 5

Another one I like for how simply it is written is as follows: $$\sum_{n=1}^{\infty}z^nH_n = \frac{-\log(1-z)}{1-z}$$ Which holds for $|z|<1$, $H_n$ being the $n$-th harmonic number $= 1 + \frac12+\frac13 \dots \frac1n$. I can't quite remember where I learned this one from.

Answer 6

You might find some interesting examples in the book, (Almost) Impossible Integrals, Sums, and Series. Here you have two examples:

First example:

$$\small\zeta(4)=\frac{4}{45}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty} \frac{(i-1)!(j-1)!(k-1)!}{(i+j+k-1)!}\left((H_{i+j+k-1}-H_{k-1})^2+H_{i+j+k-1}^{(2)}-H_{k-1}^{(2)}\right),$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1,$ denotes the $n$th generalized harmonic number of order $m$.

Second example:

Let $n\ge2$ be a natural number. Prove that $$\sum_{k_1=1}^{\infty}\left(\sum_{k_2=1}^{\infty}\left(\cdots \sum_{k_n=1}^{\infty} (-1)^{\sum_{i=1}^n k_i} \left(\log(2)-\sum_{k=1}^{\sum_{i=1}^n k_i} \frac{1}{\sum_{i=1}^n k_i +k}\right)\right)\cdots\right)$$ $$=(-1)^n\biggr(\frac{1}{2}\log(2)+\frac{1}{2^{n+1}}\log(2)+\frac{H_n}{2^{n+1}}-\sum_{i=1}^n\frac{1}{i2^{i+1}} -\frac{\pi}{2^{n+2}}\sum_{j=0}^{n-1} \frac{1}{2^j} \binom{2j}{j}$$ $$+\frac{1}{2^{n+1}}\sum_{j=1}^{n-1}\frac{1}{2^j}\binom{2j}{j}\sum_{i=1}^{j}\frac{2^i}{\displaystyle i \binom{2i}{i}}\biggr),$$ where $H_n=\sum_{k=1}^n\frac{1}{k}$ denotes the $n$th harmonic number.

Answer 7

Suppose $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are both divergent. Then, one might assume that $\sum_{n=1}^{\infty} (a_n+b_n)$ also diverges.

This is false. Suppose $a_n=1$ and $b_n=-1$ for all $n$. Then

$$\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} \,1 ~~\text{diverges}$$ and $$\sum_{n=1}^{\infty} b_n=\sum_{n=1}^{\infty} \,(-1) ~~\text{diverges}$$

However

$$\sum_{n=1}^{\infty} (a_n+b_n)=\sum_{n=1}^{\infty} \,(1+(-1)) =\sum_{n=1}^{\infty}\,0=0$$ is convergent.

Answer 8

To add another; I was surprised when I learned the two sums: $$\sum_{k=1}^{\infty}\frac1{k^2} = \frac{\pi^2}{6}$$ $$\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2} = \frac{\pi^2}{12}$$ And thought the intuition behind the second coming from the famous first sum was neat.

Answer 9

A series from user Reuns, which he proves in a previous question of mine: $$\sum_{k=1}^\infty\frac{\Re(i^{\sigma_0(k)})}{k^s} = \zeta(s)-\zeta(2s)-2\zeta(2s)\sum_{r\ge 1} (-1)^{r}\sum_{p \text{ prime}}p^{-s(2r+1)}$$ For $s>1$. (Will remove upon Reuns's request)