Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$

Question

I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$

I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.

Please help___.

Answer 1

Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .

$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$ Hope it helps!!!

Answer 2

Hint: We can write $$\frac {1}{n(n+1)} =\frac {1}{n} -\frac {1}{n+1} $$ Can you take it from here?

Answer 3

Generally, for $a_n = a_0+(n-1)d$,$\frac{1}{a_n a_{n+1}}=\frac{1}{d}(\frac{1}{a_n}-\frac{1}{a_{n+1}})$.

Similarly, higher order fractions $\frac{1}{a_n a_{n+1} a_{n+2}} =\frac{1}{2d}(\frac{1}{a_n a_{n+1}} - \frac{1}{a_{n+1}a_{n+2}})$.

When seeing these kinds of sequence series, you can have a try to split items cancel each other.