Compute $\lim_{n \to \infty} n(\frac{{\pi}^2}{6} -( \sum_{k=1}^{n} \frac{1}{k^2} ) ) $

Basel series $$ \lim_{n\to \infty} \sum_{k=1}^{n} \frac{1}{k^2} = \frac{{\pi}^2}{6} $$
is well known. I'm interested in computing the limit value $$ \lim_{n \to \infty} n\left(\frac{{\pi}^2}{6} - \sum_{k=1}^{n} \frac{1}{k^2} \right) $$

although I am not sure even whether this limit exists or not..

Does this limit exists? If exists, how can we compute this?

Thanks in advance.

By integral test,

$$\int_{n+1} ^\infty \frac{1}{x^2} dx \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \int_n ^\infty \frac{1}{x^2} dx$$

Thus

$$\frac{1}{n+1} \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \frac{1}{n}. $$

Hint

By definition, $$\sum_{k=1}^{n} \frac{1}{k^2} = H_n^{(2)}$$ where appears the harmonic number. For large values of $n$, $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ Then, the result.

By the Stolz rule, for $a_n =\frac{\pi^2}{6} - \sum\limits_{k=1}^n \frac{1}{k^2}$ and $b_n = \frac{1}{n}\downarrow 0$, $$ \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{a_{n-1}-a_n}{b_{n-1}-b_n} = \lim_{n\to\infty} \frac{\frac{1}{n^2}}{\frac{1}{n-1}-\frac{1}{n}} = 1. $$

This also can be used to get further asymptotics, e.g. $$ \lim_{n\to\infty} \frac{\frac{\pi^2}{6} - \sum\limits_{k=1}^n \frac{1}{k^2} - \frac{1}{n}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{\frac{1}{n^2} - \frac{1}{n(n-1)}}{\frac{1}{(n-1)^2} - \frac{1}{n^2}} = \frac{1}{2} $$ et cetera.