Difficulty proving $x_n$ satisfaction in the inequality I am trying to prove that $$x_n = \int_{0}^{1}\left(\frac {n}{x}\right)^n \,dx$$ satisfies the inequalities
$$\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
$$\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$

I tried to make the substitution $\frac {n}{x} =t$ but I couldn't get the desired result.
 A: $$x_n = n \int_{n}^{\infty} \frac {t^n}{t^2}\,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {t^n}{t^2} \,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {(t-k)^n}{t^2} \,dy =  n \sum_{k=n}^{\infty} \int_{0}^{1} \frac {y^n}{(k+y)^2}\,dy = n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{(k+y)^2}\right) \,dy $$
Since $\frac {1}{(k+1)^2} < \frac {1}{(k+y)^2} < \frac {1}{k^2}$, for positive integers $k$, and $y \in (0,1)$, it follows that
$$n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{(k+1)^2}\right) \,dy < x_n < n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{k^2}\right) \,dy$$
Thus, 
$$\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
Which implies that $\lim\limits_{n\to \infty} x_n = 0$. 
The second limit equals $1$. We have, based on the preceding inequalities, that
$$\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
and the result follows since $\displaystyle\lim\limits_{n \to \infty} n\left(\frac {\pi^2}{6}-\sum_{k=1}^{n} \frac {1}{k^2}\right) = 1$
