# How to evaluate $\int_1^\infty\frac{1}{x^2}\,\mathrm{d}x$ using Riemann sums?

I'm trying to integrate $\dfrac{1}{x^2}$ from $1$ to $\infty$ using Riemann sums. I know the answer is supposed to be $1$ (using the Fundamental Theorem of Calculus), but I don't know how to get there using Riemann sums.

To clarify what I mean by "using Riemann sums", I mean calculating the area of rectangles under the curve, and taking the limit as the number of rectangles approaches infinity. Apologies if I am using the term "Riemann sum" incorrectly, but that is what I mean.

I find evaluating integrals this way to be an interesting exercise.

• $$\int_1^\infty\frac1{x^2}~\mathrm dx=\lim_{R\to\infty}\int_1^R\frac1{x^2}~\mathrm dx$$Now apply Riemann sums to the RHS. – Simply Beautiful Art May 28 '17 at 11:28

That said, you can finesse things here by "partitioning" the interval $1 \leq x$ into infinitely many pieces: Fix a real number $r > 1$, consider the "infinite geometric partition" $(r^{k})_{k=0}^{\infty}$, and note that for each integer $k \geq 1$, you have $$\frac{1}{r^{2k}} = \frac{1}{(r^{k})^{2}} \leq \frac{1}{x^{2}} \leq \frac{1}{(r^{k-1})^{2}} = \frac{r^{2}}{r^{2k}}$$ on the interval $[r^{k-1}, r^{k}]$, whose length is $\Delta r_{k} = r^{k} - r^{k-1} = r^{k-1}(r - 1)$. The respective "lower" and "upper sums" are \begin{align*} L(f, P) &= \sum_{k=1}^{\infty} \frac{r^{k-1}(r - 1)}{r^{2k}} = (r - 1) \sum_{k=1}^{\infty} \frac{1}{r^{k + 1}} = (r - 1) \frac{(1/r)^{2}}{1 - (1/r)} = \frac{1}{r}, \\ U(f, P) &= \sum_{k=1}^{\infty} \frac{r^{k+1}(r - 1)}{r^{2k}} = r^{2}\, L(f, P) = r. \end{align*} Letting $r \to 1$ gives the integral. (Of course, this argument doesn't exactly pass muster in the 21st century, but it can be tightened up easily enough, and is in the spirit of what you seek.)
$f(x)=\frac{1}{x^2}$ is a convex function on $[1,+\infty)$, hence by the Hermite-Hadamard inequality it is enough to estimate the upper (or lower) Riemann sums on the $[1,M+1]$ interval. So we need to deal with:
$$\lim_{n\to +\infty}\frac{M}{n}\sum_{k=1}^{n} \frac{1}{\left(1+\frac{kM}{n}\right)^2}=\lim_{n\to +\infty}\frac{n}{M}\sum_{k=1}^{n}\frac{1}{\left(k+\frac{n}{M}\right)^2}\tag{1}$$ and if we show that such limit is $1-O\left(\frac{1}{M}\right)$ we are done, since that implies $\int_{1}^{+\infty}\frac{dx}{x^2}=1$ as wanted. We may find very tight upper and lower bounds for the sum appearing in the RHS of $(1)$ in terms of telescopic sums. For instance: $$\frac{1}{\left(k-1+\frac{n}{M}\right)\left(k+\frac{n}{M}\right)}\leq \frac{1}{\left(k+\frac{n}{M}\right)^2}\leq \frac{1}{\left(k-\frac{1}{2}+\frac{n}{M}\right)\left(k+\frac{1}{2}+\frac{n}{M}\right)}\tag{2}$$ hence it follows that: $$\sum_{k=1}^{n}\frac{1}{\left(k+\frac{n}{M}\right)^2} \geq \frac{1}{\frac{n}{M}}-\frac{1}{n+\frac{n}{M}},$$ $$\sum_{k=1}^{n}\frac{1}{\left(k+\frac{n}{M}\right)^2} \leq \frac{1}{\frac{1}{2}+\frac{n}{M}}-\frac{1}{n+\frac{1}{2}+\frac{n}{M}},\tag{3}$$ and: $$1-\frac{1}{M+1}\leq \frac{n}{M}\sum_{k=1}^{n}\frac{1}{\left(k+\frac{n}{M}\right)^2} \leq \frac{2n}{M+2n}\leq 1\tag{4}$$ so $\frac{n}{M}\sum_{k=1}^{n}\frac{1}{\left(k+\frac{n}{M}\right)^2}=1-O\left(\frac{1}{M}\right)$ follows by squeezing and we are done.
$$\lim_{X\to+\infty}\lim_{n\to+\infty}\frac {X-1}{n}\sum_{i=0}^n\frac {1}{(1+i\frac {X-1}{n})^2}$$