Integral-Summation inequality.

The equation $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ can be rewritten as $$\frac{1}{n^2}=\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$

Now, simply sum this for $n=1,2,\ldots,N$ and get: $$\begin{align}\sum_{n=1}^N\frac{1}{n^2}=&\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\\=&\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\sum_{n=1}^N{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\ln{x}\sum_{n=1}^N{\frac{1}{x^{n+1}}\:dx}\end{align}$$

(To get the inequality, we used that the integrand is positive on $(1,\infty)$, which shows that $\int_{{\rm e}^{1/n}}^{\infty}\leq\int_{{\rm e}^{1/N}}^{\infty}$ for $N\geq n$ and equality holds if and only if $N=n$.) The sum here is just the sum of a finite geometric progression. Thus, we have $$\sum_{n=1}^N{\frac{1}{x^{n+1}}}=\sum_{n=1}^N{(x^{-1})^{n+1}}=x^{-2}\sum_{n=1}^{N}{(x^{-1})^{n-1}}=x^{-2}(\frac{(x^{-1})^N-1}{x^{-1}-1})=\frac{x^{-N}-1}{x-x^2}$$

This gives us $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ which is exactly what we wanted.