Math

Computing integer sums using l’Hôpital’s rule

Posted by Dave Richeson on

Now that the busy semester is over, I’ve been able to catch up on some reading. Yesterday I read William Dunham’s article “When Euler Met l’Hôpital,” in the February 2009 issue of Mathematics Magazine. The aim of the article is to showcase some of Euler’s applications of l’Hôpital’s rule in his Institutiones calculi differentialis (1755). I thought I’d show one example here.

A few months ago I wrote about how to sum a finite number of kth powers. Here Euler presents another way of computing these sums. This is by no means the simplest way of doing so, but it is extremely clever, and it uses l’Hôpital’s rule.

Our aim is to prove the following theorem:

\displaystyle 1+2+\cdots+n=\frac{n(n+1)}{2}.

First, we assume we know sum of the first n+1 terms of a geometric series:

\displaystyle 1+x+x^2+\cdots+x^n=\frac{x-x^{n+1}}{1-x} for |x|<1.

Differentiate both sides with respect to x to obtain:

\displaystyle 1+2x+3x^2+\cdots+nx^{n-1}=\frac{1-(n+1)x^{n}+nx^{n+1}}{(1-x)^2}.

Then multiply both sides by x:

\displaystyle x+2x^2+3x^3+\cdots+nx^{n}=\frac{x-(n+1)x^{n+1}+nx^{n+2}}{(1-x)^2}.

The result follows after two applications of l’Hôpital’s rule:

\begin{array}{rl}\displaystyle 1+2+3+\cdots+n&=\displaystyle \lim_{x\to 1}(x+2x^2+3x^3+\cdots+nx^{n})\\&\displaystyle=\lim_{x\to 1}\frac{x-(n+1)x^{n+1}+nx^{n+2}}{(1-x)^2}=\frac{0}{0}\\(\text{by l'H\^{o}pital's rule})&\displaystyle=\lim_{x\to 1}\frac{1-(n+1)^2x^{n}+n(n+2)x^{n+1}}{-2(1-x)}=\frac{0}{0}\\(\text{by l'H\^{o}pital's rule})&\displaystyle=\lim_{x\to 1}\frac{-n(n+1)^2x^{n}+n(n+1)(n+2)x^{n+1}}{2}\\&\displaystyle=\frac{-n(n+1)^2+n(n+1)(n+2)}{2}\\&\displaystyle=\frac{n(n+1)}{2}\end{array}

Dunham then goes on to explain how Euler proved that \displaystyle 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6} using l’Hôpital’s rule.

He concludes by giving another of Euler’s proofs of the Basel problem, the result that made Euler famous at the beginning of his career:

\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi}{6}.

The proof illustrates Euler’s creativity and ingenuity and another use of l’Hôpital’s rule.

3 Comments

  1. misha says:
    May 27, 2009 at 1:33 am

    Why do you need l’Hôpital’s rule? You can just carry out the division of polynomials in the right-hand side.

    1. Dave Richeson says:
      May 27, 2009 at 2:44 pm

      Thanks for the comment. One thing to mention is that this isn’t presented as the best way to compute this sum. It was simply a neat application of l’Hôpital’s rule. It appeared in Euler’s textbook as one of several examples using this rule. So there may be a better way to compute the limit (without using l’Hôpital’s rule). However, I don’t see the long division argument that you allude to. Are you suggesting doing long division at the first instance of 0/0?

  2. Roberto Miguez says:
    August 11, 2009 at 9:51 am

    Great example.

    Another creative use of the geometric series sum as well.

Comments are closed.