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

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. $$\frac{1}{x^2}$$.)

• The Abel-Plana formula may work. Commented Feb 21, 2022 at 0:30

## 2 Answers

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

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

• This is shocking. +1 Commented Feb 21, 2022 at 0:45
• Very cool relationship! I assume you could basically do this with any integral and sum with the same values, right? Of course, they wouldn't be as nice, but still. Commented Feb 21, 2022 at 3:59
• Whoa amazing!! Thank you so much! Could you elaborate on why $\frac{1}{2^n} - \frac{1}{2^{n+1}}$ simplifies to $\frac{1}{2^{n+1}}$ in the second-to-last step?
– Toph
Commented Feb 21, 2022 at 22:41
• @Toph $\frac{1}{2^n}-\frac{1}{2^{n+1}}=\frac{1}{2^n}(1-\frac{1}{2})=\frac{1}{2^{n+1}}$ Commented Feb 21, 2022 at 23:01

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.

• This is a cool way of thinking about it, & I learned the term “telescoping” (en.wikipedia.org/wiki/Telescoping_series) — thank you! Trying to remember how substitution works… so like, integrating from 1 to $\infty$ turns into integrating from $\frac{1}{0}$ to $\frac{1}{1}$? Sorry I'm slow but could you write the last step where you sub in $\frac{1}{x}$ and evaluate something minus something? And then in the latter the only term you're left with is $\frac{1}{2^0}$! Nice, nice
– Toph
Commented Feb 25, 2022 at 7:41
• @Toph Right, for substitution it may help to check out wiki, including examples Commented Feb 25, 2022 at 12:23
• @Toph Note $1/x^2= u^2, dx=-du/u^2$, $x=1\implies u=1$, $x=+\infty\implies u=0.$ The negative sign is dropped when the bounds are flipped. Commented Feb 25, 2022 at 12:27