Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?

Question

Can someone give a simple explanation as to why the harmonic series

$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

doesn't converge, on the other hand it grows very slowly?

I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.

Answer 1

Let's group the terms as follows:

Group $1$ : $\displaystyle\frac11\qquad$ ($1$ term)

Group $2$ : $\displaystyle\frac12+\frac13\qquad$($2$ terms)

Group $3$ : $\displaystyle\frac14+\frac15+\frac16+\frac17\qquad$($4$ terms)

Group $4$ : $\displaystyle\frac18+\frac19+\cdots+\frac1{15}\qquad$ ($8$ terms)

$\quad\vdots$

In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $\dfrac1{2^n}$. For example all elements in group $2$ are larger than $\dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} \cdot \dfrac1{2^n} = \dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $\dfrac1{2}$, 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 $$ 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $\frac{1}{x}$. This definite integral is ln(n), so you should expect $1+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{n}$ 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 $\frac{1}{x}$, 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

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:

$$1 + \frac{1}{2} > \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$$

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

$$\frac{1}{5} + \frac{1}{6} > \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

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

Answer 4

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 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$.

Answer 5

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.

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EDIT

It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.

Answer 6

Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:

$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$

Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.

The proof is complete.

Answer 7

Let $r=\frac{m}{n}$ be any positive rational number. Beginning with the equation

$$ r=\underbrace{\frac{1}{n}+\frac{1}{n}+\dots+\frac{1}{n}}_{m\text{ times}} $$

and repeatedly applying the algebraic identity

$$ \frac{1}{p}=\frac{1}{p+1}+\frac{1}{p(p+1)} $$

we can eventually find an Egyptian fraction representation

$$r=\frac{1}{n_1}+\frac{1}{n_2}+\dots+\frac{1}{n_k}$$

of $r$ with $n_1<n_2<\dots<n_k$. So $r<H_{n_k}$.

Since $r$ was arbitrary, it follows that the sequence of harmonic numbers is unbounded. Thus the harmonic series diverges.

(I think the greedy algorithm for constructing Egyptian fractions doesn't work here, as you already need to know that the harmonic series diverges in order to prove that the greedy algorithm works for arbitrarily large rationals...)