Mean of the harmonic numbers

Posted by David Zaslavsky on November 25, 2011 2:22 AM — Edited November 25, 2011 2:22 AM

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A while ago, somebody posed an interesting problem on Physics Forums: how to evaluate the infinite sum

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k-1}{k}$$

It’s not hard to start: convert it to

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k} - \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}$$

The first term is obviously equal to 1, and in the second term, the series $\sum_{k=1}^{n}\frac{1}{k}$ is well-known under the name “harmonic numbers.” It’s easy to look up the behavior of this sum as $n\to\infty$ and thereby determine the answer, but I’m not interested in the answer. I’m interested in the method that you could use if you didn’t have the world’s mathematical references at your fingertips.

A common way to evaluate a sum with large numbers of terms like this is to approximate it by an integral. You may know that the Riemann integral, which is the first definition of an integral that students in intro calculus classes usually learn, is nothing more than the infinite limit of a sum:

$$\int_a^b f(x)\udc x = \lim_{n\to\infty}\sum_{i=0}^{n-1}f\biggl(a+i\frac{b-a}{n}\biggr)\frac{b-a}{n}$$

(technically this is only one possible way to define a Riemann integral, but that doesn’t matter here). If you flip this formula around and set $a = 1$, $b = n + 1$, and $f(x) = \frac{1}{x}$, and perform the change of iterator variable to $k = i + 1$, you wind up with

$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k} = \lim_{n\to\infty}\int_1^n\frac{1}{k}\,\udc k = \lim_{n\to\infty}\ln n$$

This in turn implies that

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k} = \lim_{n\to\infty}\frac{\ln n}{n} = 0$$

So the second term contributes nothing to the value of the original sum; evidently it’s just equal to 1.