Infinite series that surprisingly converge? [closed]

Solution 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.

Solution 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).

Solution 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.

Solution 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,

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

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.