I tried to open my bank term deposits in zeta-functions-amounts (multipled by 1000; 1644.93£, 1202.05£,...) then I realized that the sum of them approaches not only to an integer but also that integer is equal to the last one! Numerically explained : $s_{10} = \sum_2^{10} \zeta(n) =9.999014...$, $s_{50} = \sum_2^{50} \zeta(n) 49.99999999999999911... $ and so on. I tried for a while I failed to find out :
Why 1- the sum approaches to an integer, and 2- that integer is $n$ itself in $s_n$?
Notice that
$$ \sum_{n=2}^{N} \zeta(n) = \sum_{n=2}^{N} \sum_{k=1}^{\infty} \frac{1}{k^n} = N - 1 + \sum_{k=2}^{\infty} \sum_{n=2}^{N}\frac{1}{k^n} $$
and
$$ \sum_{k=2}^{\infty} \sum_{n=2}^{N}\frac{1}{k^n} = \sum_{k=2}^{\infty} \frac{1-k^{1-N}}{k(k-1)} = \underbrace{\sum_{k=2}^{\infty} \left( \frac{1}{k-1} - \frac{1}{k} \right)}_{=1} - \sum_{k=2}^{\infty} \frac{1}{k^N(k-1)}. $$
So it follows that
$$ \sum_{n=2}^{N} \zeta(n) = N - \sum_{k=2}^{\infty} \frac{1}{k^N(k-1)}, $$
where the summation part decays at most exponentially fast in $N$.