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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.

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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.

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

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