Why does the series $\frac 1 1 + \frac 12 + \frac 13 + \cdots$ not converge?

Question

Can someone give a simple explanation for why the harmonic series

$$\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

doesn't converge, but just grows very slowly?

I'd prefer an easily comprehensible explanation rather than a rigorous proof of the type I could get from an undergraduate text book.

Answer 1

Let's group the terms as follows:

group 1: 1/1 (1 term)

group 2: 1/2 + 1/3 (2 terms)

group 3: 1/4 + 1/5 + 1/6 + 1/7 (4 terms)

group 4: 1/8 + 1/9 + ... + 1/15 (8 terms)

$\vdots$

In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group n is larger than $1/2^n$. For example all elements in group 2 are larger than $1/2^2$. So the sum of the terms in each group is larger than $2^{n-1} \cdot (1/ 2^n) = 1/2$. Since there are infinitely many groups, and the sum in each group is larger than $1/2$, it follows that the total sum is infinite.

This proof is often attributed to Nicole Oresme.

Answer 2

This is not as good an answer as AgCl's, nonetheless people may find it interesting.

If you're used to calculus then you might notice that the sum 1+1/2+1/3+...+1/n is very close to the integral from 1 to n of 1/x. This definite integral is ln(n), so you should expect 1+1/2+1/3+...+1/n to grow like ln(n).

Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of ln(x) is 1/x, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.

Answer 3

There is a fantastic collection of 20 different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof 14, which appeals to triangular numbers for a sort of cameo role.

Answer 4

The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.

I also like the following argument. I'm not sure what students who are new to the topic will think about it.

Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:

1 + 1/2 > 1/2 + 1/2 = 2/2 = 1

1/3 + 1/4 > 1/4 + 1/4 = 2/4 = 1/2

1/5 + 1/6 > 1/6 + 1/6 = 2/6 = 1/3

Continuing in this way, we get S > S, a contradiction.

Answer 5

An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=\ln(1+x)$. Then $f'(x)=\dfrac {1}{1+x}$ and $ f'(0)=1$. Hence

$$\displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=\lim_{x\to 0}\dfrac{\ln(1+x)-\ln(1)}{x-0}=1,$$

and

$$ \displaystyle\lim_{n\to\infty} \dfrac{\ln\left(1+\dfrac{1}{n}\right)}{\dfrac {1}{n}}=1>0.$$

So, the series $\displaystyle\sum\dfrac{1}{n}$ and $\displaystyle\sum\ln\left(1+\dfrac {1}{n}\right)$ are both convergent or divergent. Since

$$\ln\left(1+\dfrac {1}{n}\right)=\ln\left(\dfrac{n+1}{n}\right)=\ln (n+1)-\ln(n),$$

we have

$$\displaystyle\sum_{n=1}^N\ln\left(1+\dfrac {1}{n}\right)=\ln(N+1)-\ln(1)=\ln(N+1).$$

Thus $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\dfrac {1}{n}\right)$ is divergent and so is $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$.