Using Maple to compare $\pi^2$ and the partial sums of $6\sum\_{n=0}^{\infty}\frac{1}{n^2}$ I have noticed something that appears strange. 
For instance, let $S\_{k}=6\sum\_{n=0}^{k}\frac{1}{n^2}$ be the kth partial sum, for the first 50 digits we, have


9.8636074000893588188343481429332935379126206789621 ($S\_{1000}$)
9.8696044010893586188344909998761511353136994072408 ($\pi^2$ )


Notice that the first three digits coincide (this is of course expected), but we have a strange coincidence of digits that are quite surprising.
A table with the difference between the digits is given below (read from right to left!)


[-7, 2, 2, 7, 1, 7, 2, -9, -7, 0, -1, 0, 6, 2, 0, 4, 2, 4, 1, -3, -4, 1, -7, -5, -8, 8, -5, -1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, -1, 0, 0, 3, 0, 0, -6, 0, 0, 0],
The zeros indicate that the digits coincide.


Below the tables for


$k=10^4$:
[-7, 1, 4, 7, -1, 1, -3, -4, -1, 3, 4, 1, -1, -4, -1, 3, 4, 1, 7, -5, -1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 3, 0, 0, 0, -6, 0, 0, 0, 0]


$k=10^5$:
[-4, 9, 2, 6, 1, 7, 5, -2, -7, -5, 2, 7, 5, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, -8, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 4, -1, 0, 0, 0, 0]


$k=10^6$:
[-1, 3, 4, 1, -3, 6, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, -7, 1, 0, 0, 0, 0, 4, 9, -1, 0, 0, 0, 0]


Notice that in the table above the initial (from right to left) expected sequence of zeros has length 4, but we can see a sequences with 11 and 12 consecutive zeros!
Can someone explain this surprising abundance of coinciding digits?


I have checked that a similar phenomenon occurs for other values of Riemann's $\zeta$ function. For instance:
If we compare $\zeta(3)$ and the partial sums of $\sum\_{n=1}^{\infty} \frac{1}{n^3}$ we have, for 100 digits and $k=10^{11}$ the following table


[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, -2, 7, -4, -3, -3, -3, -3, 7, 6, -4, -3, -3, 7, -4, 7, -4, -3, -3, -3, 2, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]


so we have the first 21 digits that coincide (ok, expected) and "much later" 21 that don't coincide followed by 19 that coincide.


Is this a well known phenomenon?



Strange convergence of Euler's series for $\zeta(2)$
Yes, this is a well-known occurrence of Bernoulli numbers arising from Euler-MacLaurin summation applied to the zeta function. See the AMM [article](http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P45.pdf) of Borwein-Borwein-Dilcher for details.


