Writing sums as integrals

Question

There are many proofs of the Basel problem (see this wonderful thread Different ways to prove $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$ (the Basel problem)), many of which require a step writing either $\log (1-x)$ or $1/(1-x)$ as their power series to go from the sum $\sum 1/n^2$ to an integral.

Is there a way to go from the sum to an integral without using these power series representations, but rather writing the sum directly as an integral? By "writing a sum as integral" I mean calculus-first-course-style things like $$\lim _{N\rightarrow \infty }\frac {1}{N}\sum _{n=1}^N\frac {1}{1+n/N}=\int _0^1\frac {dx}{1+x}.$$

Answer 1

If you are looking for a Riemann sum take in account: $$I=\int_{0}^{\pi/2}\left(\frac{1}{x^2}-\frac{1}{\sin^2{x}} \right)dx=-\frac{2}{\pi}\tag{1}.$$ Consider the partition: $P_{n}=\frac{\pi k}{2n}$ then $I$ can be rewritten as: $$\lim_{n}\left(\pi\sum_{k=1}^{n}\frac{1}{n(\cos{(\frac{\pi k}{n}})-1)}+\frac{2n}{\pi}\sum_{k=1}^{n}\frac{1}{k^2}\right)=-\frac{2}{\pi}\tag{2}.$$ Or which is the same: $$\frac{\pi^2}{2}\sum_{k=1}^{n}\frac{1}{n^2(1-\cos{(\frac{\pi k}{n}}))} \sim \sum_{k=1}^{n}\frac{1}{k^2} \tag{3}.$$ So to solve the Basel problem is enough to show that. $$\lim_{n}\frac{1}{n^2}\sum_{k=1}^{n}\frac{1}{(1-\cos(\frac{\pi k}{n}))}=\frac{1}{3}\tag{4}.$$ This can be solved following Spivak's calculus book.

Updated

It can be proven by induction that (check it): $$\sum_{k=1}^{n}\frac{1}{(1-\cos(\frac{\pi k}{n}))}=\frac{2n^2+1}{6}\tag{5},$$ then using $(5)$ we can conclude $(4)$ and $(3)$ which is the Basel problem.

Answer 2

One of the things you can do sometimes is: $$ \sum_{k=a}^b f(k)=\sum_{k=a}^b\int_{\sigma_1}^{\sigma_2}\bar f(x,k) dx=\int_{\sigma_1}^{\sigma_2}\sum_{k=a}^b \bar f(x,k) dx $$ Which doesn't say much at first sight, but an example:
One of the definitions of harmonic numbers: $$ H_n= \sum_{k=1}^n \frac{1}{k}= \sum_{k=1}^n \int_0^1 x^{k-1} dx=\int_0^1 \sum_{k=1}^n x^{k-1}dx=\int_0^1 \sum_{k=0}^{n-1} x^{k}dx=\int_0^1\frac{1-x^n}{1-x}dx\\\forall n\ge1 $$