How to compute the sum of a $\frac{1}{n(n+1)}$ serie?

Question

I'm solving few math puzzles to train myself for a local math contest, and I'm stuck with this problem :

Compute the sum of :

$$\frac{1}{2*1} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{2013*2014}$$

As said in the title, I know that all of those, can be written as : $\frac{1}{n(n+1)}$

But the $n(n+1)$ is not a geo seq nor an arithmetic one . So there are no theorems to compute their sum .

I'v also found a good number of similar puzzles so how can I solve this kind of problems ?

Telescoping: $$\dfrac1{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}.$$ — Workaholic, May 06 '16 at 19:38
See also these questions (they're about infinite series, but what matters is the general strategy): $\sum \limits_{r=1}^{\infty} \frac{1}{r(r+1)}$ and the related questions mentioned here, such as, $\sum\limits_{r=1}^{\infty} \frac{1}{r(r+3)}$, $\sum\limits_{r=2}^{\infty} \frac{1}{(r-1)r(r+1)}$. — Workaholic, May 06 '16 at 19:45

score 1 · Accepted Answer · answered May 06 '16 at 19:41

1

The $\frac{1}{n(n+1)} $ is called a telescopic sum, which telescopically can be expressed as : $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, which simplifies your initial expression. (This is the way that you express the series sum too).

answered May 06 '16 at 19:41

Rebellos

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How to compute the sum of a $\frac{1}{n(n+1)}$ serie?

1 Answers1