Finding a closed form of this series: $\sum_{n = 1}^{\infty} {n-\sqrt{n^2+1}\over\sqrt{n(n+1)}}$

Question

I am looking for $$\sum_{n = 1}^{\infty} {n - \sqrt{n^2 + 1}\over \sqrt{n(n+1)}}$$ I rewrite the expression and see that the terms of the series are negative

$${-1 \over n\sqrt{n(n+1)}+\sqrt{n(n+1)(n^2+1)}}$$ This is where I get stuck. I would appreciate any clues.

Answer 1

I suspect there is a typo in the given series.

One has $$ \begin{align} \sum_{n = 1}^N {n - \sqrt{n^2 \color{red}{-} 1}\over \sqrt{n(n+1)}}&=\sum_{n = 1}^N\left(\frac{n}{\sqrt{n(n+1)}}-\frac{\sqrt{n^2-1}}{\sqrt{n(n+1)}}\right) \\\\&=\sum_{n = 1}^N\left(\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n}}\right) \\\\&=\sqrt{\frac{N}{N+1}}-0 \end{align} $$ giving, as $N \to \infty$,

$$ \sum_{n = 1}^\infty {n - \sqrt{n^2 \color{red}{-} 1}\over \sqrt{n(n+1)}}=1. $$