Why do we say the harmonic series is divergent? [duplicate]

Solution 1:

The sum $\frac{1}{3}+\frac{1}{4}$ is $\gt \frac{1}{2}$.

The sum $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$ is $\gt \frac{1}{2}$. This is easy, we have $4$ terms, each $\ge \frac{1}{8}$, and all but one $\gt \frac{1}{8}$.

The sum $\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}$ is $\gt \frac{1}{2}$. We have $8$ terms, each $\ge \frac{1}{16}$, and all but one $\gt \frac{1}{16}$.

The sum $\frac{1}{17}+\frac{1}{18}+\cdots +\frac{1}{32}$ is $\gt \frac{1}{2}$.

The sum of the terms $\frac{1}{33}$ to $\frac{1}{64}$ is $\gt \frac{1}{2}$.

The sum of the terms $\frac{1}{65}$ to $\frac{1}{128}$ is $\gt \frac{1}{2}$.

And so on.

Counting in the first two terms, if we add until the term $\frac{1}{4}$ our sum is $\gt 2$. If we add up to $\frac{1}{8}$, our sum is $\gt 2.5$. Adding to $\frac{1}{16}$ puts us beyond $3$. At $32$ terms, we are beyond $3.5$. At $64$ terms, we are beyond $4$. At $256$, we are beyond $5$. At $1024$, we are beyond $6$. At $4096$, we are beyond $7$.

Painfully slow! But if we are patient enough, after a while (a long long while) we will be beyond $10$, beyond $100$, beyond $1000$. But the universe may end first.

Remarks: On a calculator, the sum is finite! Because of roundoff, after a while we are just adding $0$.

The answer dealt with the series $\sum \frac{1}{n}$. It turns out that for any positive $\epsilon$, the series $\sum \frac{1}{n^{1+\epsilon}}$ converges. We can take for example $\epsilon=0.0001$. So one can say that $\sum \frac{1}{n}$ diverges extremely reluctantly, and that close neighbours converge.

Solution 2:

What you are trying to argue is that if $a_n$ is a sequence of non-negative numbers, and $a_n\to 0$ as $n\to\infty$ then the sum $\sum_{i=1}^\infty a_n$ ought to converge.

Unfortunately, that is not true. It would simplify a lot of math if it was true.

Have you seen the proof that it diverges?

Let $$s_m=\sum_{n=1}^m \frac 1 n$$

be the $m$th partial sum.

Then $$s_{2m}-s_{m} = \sum_{n=m+1}^{2m} \frac{1}{n} \geq \sum_{n=m+1}^{2m}\frac{1}{2m} = \frac{1}{2}.$$

In particular, you can show by induction that $s_{2^k} \geq \frac{k}{2}$. So $s_m$ can be made as large as you like.

Solution 3:

The idea with the harmonic series is that you can let $$ \sum_{n=1}^N \frac 1n $$ to be as large as you want. It's not because we "cannot compute it precisely enough" that we label it infinity ; it because it doesn't make sense to attribute a real number to its value in the sense of $$ \lim_{N \to \infty} \sum_{n=1}^N \frac 1n. $$ If you have ever seen the proof of the divergence of this sum, here it is : $$ \sum_{n=1}^{2^N} \frac 1n = 1 + \sum_{i=1}^{N} \sum_{n=2^{i-1}+1}^{2^i} \frac 1n \ge \sum_{i=1}^{N} \sum_{n=2^{i-1}+1}^{2^i}\frac 1{2^i} = \sum_{i=1}^{N} \frac 12 = \frac N2 $$ and letting $N \to \infty$ means that in particular the sum $$ \sum_{n=1}^K \frac 1n \ge \sum_{n=1}^{2^N} \frac 1n \ge \frac N2 $$ if we take $K$ such that $K > 2^N$. If you don't understand the formalism, the idea is that I just regrouped the terms like this : $$ 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \dots \ge 1 + \frac 12 + (\frac 14 + \frac 14) + (\frac 18 + \frac 18 + \frac 18 + \frac 18) + \dots $$ and that by using $2^i$ terms each time, I can bound by below each term of, say $\frac 1{17},\frac 1{18}, \dots, \frac 1{32}$ by $\frac 1{32}$, and since there are $16$ such terms, the sum of them is $\frac 12$. And it keeps going on like this.

Hope that helps,