Limit of $\sum \frac1{1^2+2^2+\cdots+n^2}$ 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
 A: Assuming the formula
$$u_n=24(H_n-H_{2n+1})+\frac{6}{n+1} + 18$$
is correct, one has
$$H_n=\ln n+\gamma+O(1/n)$$ and so
$$u_n=24\ln\frac{n}{2n+1}+O(1/n)+\frac{6}{n+1} + 18$$
which means that
$$\lim_{n\to\infty}u_n=-24\ln2+18.$$
A: Since you did prove that $$u_n=24(H_n-H_{2n+1})+\dfrac{6}{n+1} + 18$$ we could even go beyond the limit itself using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12p^2}+O\left(\frac{1}{p^4}\right)$$
Use it twice and continue with Taylor series for large values of $n$ to get
$$u_n=18-24 \log (2)-\frac{3}{2n^2}+\frac{3}{n^3}+O\left(\frac{1}{n^4}\right)$$which shows the limit and also how it is approached.
I also allows a quick evaluation of $u_n$ even for small values on $n$. For example
$$u_{10}=\frac{13114793}{9699690}\approx 1.35208$$ while the above truncated series would give $\frac{4497}{250}-24 \log (2)\approx 1.35247$.
A: You already did most of the work actually. Using your $u$, we can plug in approximations for $H_n$. Using the approximation of $$H_n = \ln(n) + \gamma + O\left( \frac{1}{n} \right)$$
we can get that $$u_n \approx 24(\ln(n) + \gamma - \ln(2n+1) - \gamma) +\frac{6}{n+1}+18 = 24\ln\left( \frac{n}{2n+1} \right) + \frac{6}{n+1}+18$$
The $6/(n+1)$ term approaches $0$. Since $\frac{n}{2n+1}$ approaches $1/2$, the limit is $$24\ln(1/2) + 18 = 18-24\ln(2) \approx 1.364$$
A: $$(1/6)(\sum_{j=1}^nj^2)^{-1}=\frac {1}{n(n+1)(2n+1)}=$$ $$=(\frac {1}{n}-\frac {1}{n+1})\cdot\frac {1}{2n+1}=$$ $$=\frac {1}{n(2n+1)}-\frac {1}{(n+1)(2n+1)}=$$ $$=(\frac {1}{n}-\frac {2}{2n+1})-(\frac {-1}{n+1}+\frac {2}{2n+1})=$$ $$=\frac {1}{n}+\frac {1}{n+1}-\frac {4}{2n+1}.$$
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.
A: This will be hard to solve by hand.
By Mathematica:
$$\sum\limits_{k=1}^n \frac{6}{k (k + 1) (2 k + 1)} = 6 \left(2 H_n-2 H_{n+\frac{1}{2}}+\frac{1}{n+1}+3-4 \log (2)\right) ,$$
where $H_n$ is the Harmonic number.
