Prove the following definite integrals as limits of sum $\int_a^b \frac{1}{x^2}dx = 1/a - 1/b$

Question

My attempt so far ..

$\int_a^b f(x)dx = \lim \sum_{1}^{n} h f(a+rh) $, where n tends to infinity, $nh = b-a$

Here $f(x) = \frac{1}{x^2}$ , using this we get

$\int_a^b f(x)dx = \lim \sum_{r=1}^{n^2} h f(a+rh) = \lim h [\frac{1}{(a+h)^2}+\frac{1}{(a+2h)^2}+...+\frac{1}{(a+nh)^2}]$

I cannot proceed further. can someone help?