Calculate $\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N$ How to calculate the limit
$$\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N?$$
By using the numerical method with Python, I guess the right answer is $-1$ but how to prove? I have no idea.


Let $N\geq 1$. Note that
$$
\frac{{\pi ^2 }}{6} - \sum\limits_{n = 1}^N {\frac{1}{{n^2 }}}  = \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  = \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n(n - 1)}}}  - \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 (n - 1)}}}  = \frac{1}{N} - \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 (n - 1)}}} .
$$
Then
$$
\sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 (n - 1)}}}  \le \sum\limits_{n = N}^\infty  {\frac{1}{{n^3 }}}  \le \frac{1}{{N^{3/2} }}\sum\limits_{n = N}^\infty  {\frac{1}{{n^{3/2} }}}  \le \frac{{\zeta (3/2)}}{{N^{3/2} }}.
$$
Accordingly,
$$
\left( {\sum\limits_{n = 1}^N {\frac{1}{{n^2 }}}  - \frac{{\pi ^2 }}{6}} \right)N =  - 1 + \mathcal{O}\!\left( {\frac{1}{{N^{1/2} }}} \right),
$$
showing that the limit is indeed $-1$.
