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.
 A: 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 $\approx 0.550967$.
A: Assuming we know $\lim_n \sum_{k=1}^n\frac{1}{k^2} = \frac{\pi^2}{6}$ (see e.g. https://en.wikipedia.org/wiki/Basel_problem) then
$$\lim_n  \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2} =1.$$
Now write
$$\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}=1+\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1 = 1+\frac{1}{\left(\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1\right)^{-1}}.$$
Denote $a_n := \left(\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1\right)^{-1}$. Observe that $\lim_n a_n=\infty$. Hence,
$$\lim_n \left(1+\frac{1}{a_n}\right)^{a_n} =: e,$$
by definition.
Thus,
$$\left(\frac{6}{\pi^2}\sum_{k=1}^n \frac{1}{k^2}\right)^n =\left[\left( 1+\frac{1}{a_n}\right)^{a_n}\right]^{n\ a_n^{-1}} .$$
Taking limit
$$\lim_n\left(\frac{6}{\pi^2}\sum_{k=1}^n \frac{1}{k^2}\right)^n = e^{\lim_n n \ a_n^{-1}},$$
where in the last step we used the continuity of the exponential function.
Now, let us compute $n\ a_n^{-1}$.
$$\lim_n \frac{\frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1}{\frac{1}{n}}.$$
The above limit, which is indefinite, can be computed easily by using Stolz-Cesàro Theorem (see e.g. https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem). If $c_n := \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}-1$ and $b_n:= \frac{1}{n}$ then
$$\lim_n = \frac{c_n}{b_n} = \lim_n \frac{c_{n+1}-c_n}{b_{n+1}-b_n} = \frac{\frac{6}{\pi^2}\frac{1}{(n+1)^2}}{\frac{-1}{n(n+1)}}=-\frac{6}{\pi^2}.$$
Altogether,
$$\lim_n \left( \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}\right)^n = e^{-\frac{6}{\pi^2}}.$$
A: Doesn't seem too hard. $$\frac6{\pi^2}\sum_{k=1}^n\frac1{k^2}
=1-\frac{6}{\pi^2}\sum_{k=n+1}^\infty\frac1{k^2}$$and $$\frac{6}{\pi^2}\sum_{k=n+1}^\infty\frac1{k^2}\sim\frac6{\pi^2n},$$so the limit is $$e^{-6/\pi^2}.$$

Details for anyone who thinks there are some missing: We know that $$\left(1-\frac{c}{n}\right)^n\to e^{-c}$$for $c\in\Bbb R$. This convergence is uniform on compact subsets of $\Bbb R$, and hence if $c_n\to c$ it follows that$$\left(1-\frac{c_n}{n}\right)^n\to e^{-c}$$
