Explicit formula for $S_n=\sum_{r=1}^n\dfrac1{r(r+1)}$.

Question

I was asked to find an explicit formula for $$S_n=\sum_{r=1}^n\dfrac1{r(r+1)}$$ and then go on to find the limit.

I deduced that it would give $S_n=\frac1n-\frac1{n+1},$ however I was wrong and the actual answer is $S_n=1-\frac1{n+1}$.

I have spent a while looking at it, but cannot figure out what makes my answer wrong and the other answer right.

Many thanks in advance for any help- it is much appreciated - a severely struggling into to real analysis student.

Does this answer your question? How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? — PinkyWay, Nov 10 '21 at 22:31

score 3 · Accepted Answer · answered Nov 10 '21 at 22:21

$$S_n = \sum \limits_{r=1}^n \frac1{r(r+1)} \\= \sum \limits_1^n \left(\frac1r - \frac1{r+1}\right) \\= \left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots + \left(\frac1{n-1}-\frac1n\right)+\left(\frac1n -\frac1{n+1}\right) \\= 1 - \frac1{n+1}$$

Your answer may have failed to sum the earlier terms

Explicit formula for $S_n=\sum_{r=1}^n\dfrac1{r(r+1)}$.

1 Answers1