Question

How to find this series:

Answer 1

series from 1 to infinity of 1/((n+1)(n+2))

Answer 2

hint:
we can write:
\[\Large \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}\]

Answer 3

so, we have:
\[ \begin{gathered}
\sum\limits_{n = 1}^m {\frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} = \sum\limits_{n = 1}^m {\left( {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right)} = \hfill \\
\hfill \\
= \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{5}} \right) + ... - \frac{1}{{m + 1}} + \frac{1}{{m + 1}} - \frac{1}{{m + 2}} = \hfill \\
\hfill \\
= \frac{1}{2} - \frac{1}{{m + 2}} \hfill \\
\end{gathered} \]
now what is the limit value of that sum when as m goes to +infinity?

Answer 4

Telescoping series ftw! the limit approaches 1/2! Awesome ty