Partial fraction expansion

[sasa]

One of the telescoping series produce following sum:

\[\sum_{n=1}^{\infty} \frac {1}{n(n+1)} = 1\]

In order to prove it, expression \(\frac{1}{n(n+1)}\) should be shown in form \(\frac{a}{n} + \frac{b}{n+1}\) which, hence, in this case is \(\frac{1}{n} - \frac{1}{n+1}\) and then trivial to prove.

Wolfram alpha web engine have appropriate command which gives exact expression: Partial fraction 1/(n(n+1)). Other commands including fraction, expand or similar returns different expression.

It would be interesting to show is it any modern HP calculator capable to expand expression appropriately and at end calculate the sum.