# Why does the harmonic series diverge?

My Precalculus students have been asking me this question. I don’t really have a great answer, except that it’s true. Granted it’s not very intuitive. But nothing about infinite series is intuitive. For those not in my class or not familiar with the harmonic series, the question is:

Does $\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ converge?

And the answer is no.  This is surprising, because the terms of the series approach zero.

The proof that it diverges is due to Nicole Oresme and is fairly simple. It can be found here. There are at least 20 proofs of the fact, according to this article by Kifowit and Stamps.

Interestingly, the alternating harmonic series does converge:

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln{2}$

And so does the p-harmonic series with p>1. For instance:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$

Besides looking at the sequence of partial sums, I’m not sure I can help you with the intuition on any of these facts. Can you?

## 11 thoughts on “Why does the harmonic series diverge?”

1. This is over a year old, so hope you will still read this comment…

There is a well-known way to show that the harmonic series diverges. I use it in my precalculus class and it is very effective.

let S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + … + 1/16 + 1/17 + …
then we can define a “smaller S”, call it “S_small” or just Sm, so
Sm = 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + … + 1/16 + 1/32 + …

Well, Sm is clearly less than S. And Sm quite clearly diverges, as 1/2 is being added an infinite number of times and any arithmetic series diverges. So S diverges. It’s kind of a baby comparison test.

Also, (and this I came up with myself), this way of representing S also reveals a more fundamental fact: the sum increases logarithmically because you must take an exponentially increasing number of terms to grab another constant amount of sum (in this case 1/2). So that also kind of shows why the sum from k=1 to n of 1/k is about log(n).

I love your blog! (Just discovered it today).

Will

• Will, thanks for your comment! I have seen the comparison test proof (it’s one of the proofs in the paper I linked to, and the Wikipedia article uses it too). I hadn’t ever thought of the logarithmic growth of the series, though. That’s awesome! Thanks for the great insight!

2. One can prove it using the integral test. Where integral (1/x) is equal to the sum of the series 1/n.
The integral works out to be ln x and on substituting the limits (1 and infinity), you’ll get infinity. This clearly shows that the integral diverges and hence the series diverges.Moreover,all series with ‘p’>1 will converge (again proved by integral test) . The alternating harmonic series does converge. It can be proved using Leibniz theorem(For Alternating series).Any alternating harmonic series which converges absolutely is convergent.(i.e. the alternating harmonic series n to the power of -2 is convergent as the series (modulus of the above) n to the power -2 is convergent. The reverse isn’t true! Take 1/n as an example. The alternating series converges to ln2 but the absolute series diverges!!
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• My Precalculus students were asking me this question, so I couldn’t use the integral test. Though the integral test still doesn’t stop it from being hard to swallow. The integral test IS mentioned in the article I pointed to.

And yes, the fact that the alternating series DOES converge is quite amazing!

• Thanks for stopping by, Lana. Can you explain what you mean?

Zeno’s paradox proposes we sum $1/2+1/4+1/8+\cdots$ which converges to 1 (or 2 if you start with 1 instead of 1/2). That’s a convergent series, but the harmonic series diverges, so I’m not sure I get your meaning. What am I missing?

3. Use integral test because series is the partial sums of sequence. We know the antiderivative of 1/n is ln(n), where n is approaching to infinity. So harmonic series diverges.

• Just for clarity, because integral diverges and series (start at n=1, overestimate) is greater than its integral, so harmonic series must diverge (this method is to compare series to its integral).