Find the sum of $\sum_{k=1}^n\left(\frac{1}{(k+1)^2}-\frac{1}{k^2}\right)$

Question

Hi how do I fin the sum of

$$\sum_{k=1}^n\left(\frac{1}{(k+1)^2}-\dfrac1{k^2}\right)$$

I tried everything and I can't get around it..

Answer 1

Hint : the mentioned sum is $\frac{1}{2^2}- 1 + \frac{1}{3^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{3^2} + \dots$. Do you see that you some members vanishes?

Answer 2

Expand the sum. All the terms get cancelled except two terms $(1/(n+1)^2) -1$ and this is the answer.

Answer 3

Whenever you have a sum of this form $$S(n)=\sum\limits_{i=0}^n\bigg(f(i+1)-f(i)\bigg)$$

We call it a telescopic series and only the terms with extremal indices will not vanish. Indeed $$\require{cancel}S(n)=\big[\cancel{f(1)}-f(0)\big]+\big[\cancel{f(2)}-\cancel{f(1)}\big]+\cdots+\big[f(n+1)-\cancel{f(n)}\big]=f(n+1)-f(0)$$