# Integral $\int\limits_a^b\frac{\mathrm{d}x}{x^2}$ using Riemann sums

Evaluate $$\displaystyle \int\limits_a^b\frac{\mathrm{d}x}{x^2},$$ where $$0, using Riemann sums.

Attempt. Since $$1/x^2$$ is decreasing, it is integrable and using Riemann sums we get: $$\int\limits_a^b\frac{\mathrm{d}x}{x^2}=\lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^n\frac{1}{\big(a+k\,\frac{b-a}{n}\big)^2}.$$ Is it possible to get a formula for the above sum?

This is tricky: you may approximate $$n(b-a) \sum_{k=1}^{n}\frac{1}{(na+k(b-a))^2}$$ with $$n(b-a) \sum_{k=1}^{n}\frac{1}{(na+k(b-a))(na+(k+1)(b-a))}$$ which is a telescopic sum, equal to $$n\sum_{k=1}^{n}\left[\frac{1}{na+k(b-a)}-\frac{1}{na+(k+1)(b-a)}\right]=n\left[\frac{1}{na+(b-a)}-\frac{1}{(n+1)b-a}\right]$$ whose limit as $$n\to +\infty$$ is $$\frac{1}{a}-\frac{1}{b}$$. You may check that the difference between the actual Riemann sum and its telescopic approximation is $$O\left(\frac{1}{n}\right)$$, hence we have just proved $$\int_{a}^{b}\frac{dx}{x^2} = \frac{1}{a}-\frac{1}{b}$$ (for $$b>a>0$$) by creative telescoping.