What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$

Question

How can I find the formula for the following equation?

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$

More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:

$\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.

So how should I approach finding the formula? Thanks!

Answer 1

If you simplify your partial sums, you get $\frac12,\frac23,\frac34,\frac45,....$ Does this give you any ideas?

Answer 2

Hint: Use the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and find $S_n=\sum_1^n\left(\frac{1}{k}-\frac{1}{k+1}\right)$.

Answer 3

You can use these formulas for Problems like this , $$\frac{1}{a.b}=\frac{1}{(b-a)}\left(\frac{1}{a}-\frac{1}{b}\right)$$ $$\frac{1}{a.b.c}=\frac{1}{(c-a)}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$

Hope it'll help you .