Limit of the nth power of certain partial sums Evaluate
$$\lim_{n\to \infty}\left (\frac{6}{\pi^2}\sum_{k=1}^{n} \frac{1}{k^2} \right )^n.$$
The context is: I just thought it up and thought some members of MSE would like to try it.

Consider $$A_n=\left (\frac{6}{\pi^2}\sum_{k=1}^{n} \frac{1}{k^2} \right )^n$$ Now, using harmonic numbers $$\sum_{k=1}^{n} \frac{1}{k^2} =H_n^{(2)}$$ Taking logarithms $$\log(A_n)=n\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)$$ Now, using asymptotics $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2
   n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{6}{\pi^2}H_n^{(2)}=1-\frac{6}{\pi ^2 n}+\frac{3}{\pi ^2 n^2}+O\left(\frac{1}{n^3}\right)$$ Using Taylor $$\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)=-\frac{6}{\pi ^2 n}+\frac{3 \left(\pi ^2-6\right)}{\pi ^4
   n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A_n)=n\log\Big(\frac{6}{\pi^2}H_n^{(2)} \Big)=-\frac{6}{\pi ^2}+\frac{3 \left(\pi ^2-6\right)}{\pi ^4
   n}+O\left(\frac{1}{n^2}\right)$$ Now, using $A_n=e^{\log(A_n)}$ and Taylor again $$A_n=e^{-\frac{6}{\pi ^2}}+\frac{3 e^{-\frac{6}{\pi ^2}} \left(\pi ^2-6\right)}{\pi ^4
   n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
Just for illustration purposes, using $n=10$ would lead to $A_{10}\approx 0.551038$ while the above approximation gives $\approx 0.550967$.
