Find $\sum_{r=n+1}^{\infty} \left(\frac{1}{r}-\frac{1}{r+1}\right)$

Question

Find $$\sum_{r=n+1}^{\infty} \left(\frac{1}{r}-\frac{1}{r+1}\right)=\sum_{r=1}^{\infty} \left(\frac{1}{r}-\frac{1}{r+1}\right)-\sum_{r=1}^{n} \left(\frac{1}{r}-\frac{1}{r+1}\right)\;.$$

I found that: $$\sum_{r=1}^{n}\left( \frac{1}{r}-\frac{1}{r+1}\right)=1-\frac{1}{n+1}$$ however, I am not sure how to proceed. Any ideas? Many thanks

Take the sum from $r=n$ to $r=N$ and let $N\rightarrow +\infty$ — Tuvasbien, Apr 02 '20 at 14:15
With the result for the finite sum you are almost finished. Take the limit now — B.Swan, Apr 02 '20 at 14:20
@77and33is100: The limit of the finite sum is not what’s wanted here. — Brian M. Scott, Apr 02 '20 at 14:31

score 1 · Answer 1 · answered Apr 02 '20 at 15:11

You have that $$ \begin{gathered} r_n = \sum\limits_{r = n + 1}^{ + \infty } {\left( {\frac{1} {r} - \frac{1} {{r + 1}}} \right)} = \sum\limits_{k = 1}^\infty {\left( {\frac{1} {r} - \frac{1} {{r + 1}}} \right)} - \sum\limits_{k = 1}^n {\left( {\frac{1} {r} - \frac{1} {{r + 1}}} \right)} \hfill \\ \hfill \\ = 1 - \left( {1 - \frac{1} {{n + 1}}} \right) = \frac{1} {{n + 1}} \hfill \\ \end{gathered} $$ Moreover $$ \mathop {\lim }\limits_{n \to + \infty } r_n = \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{r = n + 1}^{ + \infty } {\left( {\frac{1} {r} - \frac{1} {{r + 1}}} \right)} = \mathop {\lim }\limits_{n \to + \infty } \frac{1} {{n + 1}} = 0 $$ as it must be with the $n-th$ remainder of a convergent series as $n \to +\infty$

Find $\sum_{r=n+1}^{\infty} \left(\frac{1}{r}-\frac{1}{r+1}\right)$

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