Evaluate $\displaystyle \int\limits_a^b\frac{\mathrm{d}x}{x^2},$ where $0<a<b$, 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?

Thanks in advance.


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.

  • $\begingroup$ The approximation by a telescopic sum was the key. Great approach. $\endgroup$ – Nikolaos Skout Oct 4 '19 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.