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.

[Carsen]

The 50g can solve this in no time. It took my hardware 50g .43 seconds to get the partial fraction answer & it took .89 seconds to get the answer for the summation. I am in ALG mode to show that the 50g actually solved the problem.

And of course, the Prime does the same. A fantastic calculator.

[sasa]

Thank you, Carsen.

I believe with a finite limit (10 000, for instance) elapsed time may be much larger, as with finding limes to infinity is fairly straightforward to obtain. Anyway, correct symbolic evaluation is not nearly simple and performances are indeed respectable.