Prove the identity $ \sum_{n=0}^{N} \frac{1}{(n+1)(n+2)} = \frac{N+1}{N+2} $

Question

I have the identity $$ \sum_{n=0}^{N} \frac{1}{(n+1)(n+2)} = \frac{N+1}{N+2} $$ that I am supposed to prove by using the partial fraction decomposition $$ \frac{1}{n+1}- \frac{1}{n+2} $$ but I'm not sure how to proceed

Have you written down the first three or four terms using the PF? What do you notice? — ancient mathematician, Jan 08 '21 at 16:04

score 0 · Accepted Answer · answered Jan 08 '21 at 16:10

0

Notice that this is a telescopic sum, meaning that: $$ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{n+1} - \frac{1}{n+2} = 1 - \frac{1}{n+2} = \frac{n+2-1}{n+2}= \frac{n+1}{n+2} $$

answered Jan 08 '21 at 16:10

Contrakrostipunktus

77

Prove the identity $ \sum_{n=0}^{N} \frac{1}{(n+1)(n+2)} = \frac{N+1}{N+2} $

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