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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.

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  • $\begingroup$ What is "except those in a measure zero subset" supposed to mean here? Surely any $x \in \mathbb{R}$ is in a measure zero subset $\{x\}$. $\endgroup$ – Alex Provost Dec 5 '19 at 17:38
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    $\begingroup$ @AlexProvost I mean that the limit converges to K for almost all $x \in \mathbb{R}$, those for which it does not belonging to a measure zero set. You can read about it here: mathworld.wolfram.com/KhinchinsConstant.html $\endgroup$ – The Wheel is Before Descartes Dec 5 '19 at 17:48
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    $\begingroup$ As interesting as this is, I'm not sure this is a "good question" for this site. It's subjective and, while there are good subjective questions, this doesn't meet the critieria, in particular number 6. I'm not saying I don't find it cool, I've watched plenty of Numberphile videos that would hit the same dopamine button this question does, I'm just saying it doesn't seem a good fit $\endgroup$ – corsiKa Dec 6 '19 at 2:30
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    $\begingroup$ If you are in a Zeno-like mood, all convergent infinite series converge strangely. $\endgroup$ – John Coleman Dec 6 '19 at 3:37
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    $\begingroup$ The identity from this question is pretty peculiar at first. $\endgroup$ – Clement C. Dec 6 '19 at 6:50
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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.

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    $\begingroup$ This one is surprisingly intuitive, which makes it even better! Imagine you are summing up the harmonic series up to 1,000 and decide to take out all numbers that contain 9. Then, you are removing all numbers 9xx (900-999) plus all other numbers that contain 9. That's 1/10 of all the numbers plus all other numbers that contain 9. The higher you sum up, the more numbers you are deleting, which eventually becomes "almost all the numbers". This same philosophy applies to any sequence of numbers; by throwing them out, you are essentially throwing out "almost all the numbers". $\endgroup$ – Ty Jensen Dec 7 '19 at 2:53
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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).

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  • $\begingroup$ What is this pattern called? Does this pattern of adding iterated logs $\to$ divergent sum hold generally? $\endgroup$ – Zach466920 Dec 7 '19 at 18:54
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    $\begingroup$ @Zach466920 Yes, it does. This is called the "generalized Bertrand series", and can be proven using the Cauchy condensation test. $\endgroup$ – Clement C. Dec 7 '19 at 18:59
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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.

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    $\begingroup$ This follows from a more general fact: if $F(z) = \sum_{n \geq 0} a_n z^n$ then $$\frac{F(z)}{1 - z} = \sum_{n \geq 0} \left(\sum_{j = 0}^n a_j \right)z^n\,.$$ $\endgroup$ – Marcus M Dec 5 '19 at 19:56
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    $\begingroup$ I wanted to add that this is called the generating function for the sequence $H_n$ of harmonic numbers. $\endgroup$ – zhantyzgz Dec 5 '19 at 21:16
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    $\begingroup$ This is not only educational but an excellent example. Thanks! $\endgroup$ – Ty Jensen Dec 7 '19 at 2:56
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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,

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


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.

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  • $\begingroup$ Is there a reference for this series? It should be possible to prove this using a contour integral for the function $\frac{\sin(z)+z \cos(z)}{z^2(z\sin(z)-1)}$ over the imaginary axis and a half-circle in $\Re(z)>0$. $\endgroup$ – Thijs Dec 11 '19 at 20:03
  • $\begingroup$ @Thijs I don't know of one, I found the sum myself. I used an integral over a full circle, since the residues at the negative roots are the same as at the positive roots. $\endgroup$ – Ben Dec 11 '19 at 22:05
  • $\begingroup$ @Thijs I've updated my post to include a more general result that I also found, the generating function of $\sum x_n^{-2k}$. $\endgroup$ – Ben Dec 11 '19 at 22:20
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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.

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  • $\begingroup$ Yes, that is...quite a strange sum! $\endgroup$ – The Wheel is Before Descartes Dec 5 '19 at 23:59
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    $\begingroup$ @heepo I added another beautiful example from the same book. $\endgroup$ – user97357329 Dec 6 '19 at 0:04
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    $\begingroup$ It reminds me of the proof of QR. $\endgroup$ – The Wheel is Before Descartes Dec 6 '19 at 0:09
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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.

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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.

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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.

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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)

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