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