Harmonic Series Diverges

The harmonic series

Ever wonder whether the series $1 + \frac{1}{2} + \frac{1}{3} + ...$ diverges? (i.e. sum to infinity)

Let $H_n = 1 + \frac{1}{2} + ... + \frac{1}{n}$.

$H_{10}$ is approximately $2.9$.

$H_{100}$ is approximately $5.2$.

$H_{10000000}$ is approximately $16.7$.

Wow, this series sure is growing really slow. Seems like it would converge, or does it???

We'll in fact, prove that the harmonic series diverge. Note that

$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{2}$.

and so on. This implies that

$H_{2^n} \ge 1 + \frac{n}{2}$, so the sum to infinity diverges.

The harmonic series is the series in which the terms are the reciprocals of the natural numbers, in order: $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... = \sum_{n=1}^{\infty}\frac{1}{n}$$ It is not obvious at all what this series adds up to, or whether it adds up to anything at all. Perhaps someone who has recently seen the series: $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = \sum_{n=1}^{\infty}\frac{1}{2^n} = 1$$ may be inclined to believe that the harmonic series converges, since it also has terms that approach zero.

If you start adding up the terms, they seem to defy easy resolution. The series will increase painfully slowly, but if you persist, hoping to see a wall that the partial sums will not cross, you will see that it is not clear that the series will ever stop increasing.

So does the harmonic series converge or diverge? Let's find out.

It turns out that the harmonic series diverges. To prove this, we observe that we can always make a certain collection of terms "large enough": $$\frac{1}{1} \ge \frac{1}{2}$$ $$\frac{1}{2} \ge \frac{1}{2}$$ $$\frac{1}{3} + \frac{1}{4} \ge 2\times\frac{1}{4} = \frac{1}{2}$$ $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \ge 4\times\frac{1}{8} = \frac{1}{2}$$ $$...$$ $$\frac{1}{2^{n-1}+1} + \frac{1}{2^{n-1}+2} + ... + \frac{1}{2^n} \ge 2^{n-1}\times\frac{1}{2^n} = \frac{1}{2}$$ $$...$$ We can separate the harmonic series into groups of terms based on the powers of $2,$ as shown above, and each group has a sum that is at least $\frac{1}{2}.$ Therefore, $$\sum_{n=1}^{\infty}\frac{1}{n} \ge \sum_{n=1}^{\infty}\frac{1}{2}$$ But the second series clearly diverges! So the first series must diverge as well, and we are done.

If you are more calculus-minded, here is a great proof that the harmonic series diverges. The key is to observe the following easy-to-see truth: $$\frac{1}{n} \gt \int_{n}^{n+1} \frac{dx}{x}$$ It follows that: $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \gt \int_{1}^{2} \frac{dx}{x} + \int_{2}^{3} \frac{dx}{x} + \int_{3}^{4} \frac{dx}{x} + ... + \int_{n}^{n+1} \frac{dx}{x}$$ $$\sum_{k=1}^{n}\frac{1}{k} \gt \int_{1}^{n+1}\frac{dx}{x}$$ In the limiting case as $n$ grows without bound, $$\sum_{n=1}^{\infty}\frac{1}{n} \gt \lim_{n\rightarrow\infty}\int_{1}^{n+1}\frac{dx}{x}$$ Now, we evaluate the limit on the right side: $$\lim_{n\rightarrow\infty}\int_{1}^{n+1}\frac{dx}{x} = \lim_{n\rightarrow\infty} \left[\ln x\right]^{n+1}_{1}$$ $$\lim_{n\rightarrow\infty}\int_{1}^{n+1}\frac{dx}{x} = \lim_{n\rightarrow\infty}\ln\left(n+1\right)$$ Clearly, this diverges, since $ln x$ grows without bound as $x$ grows without bound. Thus, the limit of the integral diverges. And since the partial sum is always larger than the integral, the limit of the partial sum (that is, the infinite series) must diverge as well!

This classic example motivates a theorem called the Integral Test, which gives information about whether an infinite series converges or diverges. It is not proved here, but it is a relatively simple consequence of the Comparison Test. It states the following:

Let $f$ be a continuous, real-valued function that is defined and monotone decreasing on the interval $[N, \infty).$ Then the infinite series $$\sum_{n=N}^{\infty}f(n)$$ converges to a finite value if and only if the indefinite integral $$\int_{N}^{\infty}f(x)dx$$ converges to a finite value.