TITLE: Limit of $\sum \frac1{1^2+2^2+\cdots+n^2}$
QUESTION [6 upvotes]: it is asked to compute the limit of the sequence 
$$ u_n :=\displaystyle\sum_{k=1}^n \dfrac{1}{1^2+2^2+\cdots+k^2}$$
I used the following 
$$1^2+2^2+\cdots+k^2=\dfrac{k(k+1)(2k+1)}{6}$$
to prove that 
$$u_n=24(H_n-H_{2n+1})+\dfrac{6}{n+1} + 18$$
where $(H_n)$ is the harmonic series.
I don't if this is of any use to find the desired limit.
Thanks

REPLY [2 votes]: $$(1/6)(\sum_{j=1}^nj^2)^{-1}=\frac {1}{n(n+1)(2n+1)}=$$ $$=(\frac {1}{n}-\frac {1}{n+1})\cdot\frac {1}{2n+1}=$$ $$=\frac {1}{n(2n+1)}-\frac {1}{(n+1)(2n+1)}=$$ $$=(\frac {1}{n}-\frac {2}{2n+1})-(\frac {-1}{n+1}+\frac {2}{2n+1})=$$ $$=\frac {1}{n}+\frac {1}{n+1}-\frac {4}{2n+1}.$$
With $H_m=\sum_{n=1}^m\frac {1}{n}$ and with $\gamma$ being Euler's Constant we have
$H_m=\gamma +\ln (m)+o(1)$ as $n\to \infty,$ and
$\sum_{n=1}^{m}\frac {1}{n+1}=-1+\gamma+\ln (m+1) +o(1)=-1+\gamma+\ln (m)+o(1),$ and
$$\sum_{n=1}^m\frac {4}{2n+1}=4[-1+H_{2m+1}-(1/2)H_m]=$$ $$=4[-1+\ln (2m+1)+\gamma +o(1)-(1/2)(\gamma+\ln (m)+o(1)]=$$ $$=4[-1+\ln (2m)+\gamma -(1/2)(\gamma +\ln (m)]+o(1).$$
I leave the rest to the reader.