Is there a relationship between $\sum _{n=1}^{\infty }\left({\frac {1}{2}}\right)^{n} = 1$ and $\int_{1}^{\infty} \frac{1}{x^2} \,dx = 1$?

Question

A classic example of an infinite series that converges is:

${\displaystyle {\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots =\sum _{n=1}^{\infty }\left({\frac {1}{2}}\right)^{n}=1.}$

A classic example of an infinite integral that converges is:

$\displaystyle\int_{1}^{\infty} \frac{1}{x^2} \,dx = 1.$

They feel very similar! But not quite the same. Is there a way to think about one in terms of the other?

I ask partly because I want to borrow the nice geometric illustrations that the former converges (like this, or similarly for other geometric series) to show the latter converging. (Related question about illustrating the geometry of $\frac{1}{x}$ vs. $\frac{1}{x^2}$.)

Answer 1

You can break the integral $\int_{1}^{\infty}\frac{1}{x^2}dx$ into a sum of integrals over intervals $[2^n,2^{n+1}]$:

$$\int_{1}^{\infty}\frac{1}{x^2}dx = \sum_{n=0}^{\infty}\int_{2^n}^{2^{n+1}}\frac{1}{x^2}dx = \sum_{n=0}^{\infty} \left(\frac{1}{2^n} - \frac{1}{2^{n+1}} \right) =\sum_{n=0}^{\infty} \frac{1}{2^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{2^{n}}.$$

Answer 2

Note that by the substitution $u=1/x$, we have the integral

$$\int_1^\infty \frac{1}{x^2} dx=\int_0^1 1 du.$$

Meanwhile, we have the series $$\sum_{n=1}^\infty \frac{1}{2^n}=\sum_{n=0}^\infty \frac{1}{2^{n}}-\frac{1}{2^{n+1}}.$$

Both express the area of a unit block, the former by integrating, the latter by telescoping.