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

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 2

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

Answer 3

Observe that $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$

Answer 4

While exploitation of the resulting telescoping series after partial fraction expansion is a very simple way forward, I thought it might be instructive to present another way forward. Here, we write

$$\begin{align} \sum_{k=1}^N \frac{1}{k(k+1)}&=\sum_{k=1}^N \int_0^1y^{k-1}\,dy \int_0^1 x^k\,dx\\\\ &=\int_0^1\int_0^1x\sum_{k=1}^N (xy)^{k-1}\,dx\\\\ &=\int_0^1\int_0^1 x\frac{1-(xy)^N}{1-xy}\,dx\,dy\\\\ &=\int_0^1\int_0^x \frac{1-y^N}{1-y}\,dy\,dx\\\\ &=\int_0^1\int_y^1\frac{1-y^N}{1-y}\,dx\,dy\\\\ &=\int_0^1(1-y^N)\,dy\\\\ &=1-\frac1{N+1} \end{align}$$

as expected!

Answer 5

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 .