Limit of the nth power of certain partial sums

Solution 1:

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$$\left(1-\frac{c_n}{n}\right)^n\to e^{-c}$$

Solution 2:

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 $\approx 0.550967$.