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.

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,
$$\lim_n \left( \frac{6}{\pi^2}\sum_{k=1}^n\frac{1}{k^2}\right)^n = e^{-\frac{6}{\pi^2}}.$$
