Divergence of the infinite sum of $\frac1n$

Question

I'm asked to show that the harmonic series $$\sum_{n=1}^{\infty}\cfrac1n=1+\cfrac12+\cfrac13+\cfrac14+...$$ is divergent.

I figured it would be satisfactory enough to note that $n>1$ for all $n>1$, and so the limit of $\cfrac1n$ will approach $0$ as $n\rightarrow\infty$, then so too should its sum converge. Instead, my textbook lists out the terms $s_8, s_{16}, s_{32}$, and so on. But I have no clue what they're doing beyond $s_2$; \begin{align*} s_4&=1+\frac12+\biggr(\frac13+\frac14\biggr)>\frac12+\biggr(\frac14+\frac14\biggr)=1+\frac22 \\ s_8 &=1+\frac12+\biggr(\frac13+\frac14\biggr)+\biggr(\frac15+\frac16+\frac17+\frac18\biggr)>1+\frac12+\biggr(\frac14+\frac14\biggr)+\biggr(\frac18+\frac18+\frac18+\frac18\biggr)=1+\frac12+\frac12+\frac12=1+\frac32\end{align*}

And this continues infinitely. I don't understand exactly what's going on, especially the portions that repeat $1/n$ at the same value $n$ inside the parentheses.

Can someone explain what's happening?

Answer 1

For a full explanation:

Just because $x_n → 0$ as $n→\infty$ doesn't mean that the sum converges. The inverse is true however: if $x_n$ didn't go to $0$ as $n → \infty $ we would know the sum diverges.

Now to explain what is going on, the proof groups the sum into groups with a lower bound of $\frac{1}{2}$. If does so by making an $n$-th grouping as follows:

\begin{equation*} \begin{split} \sum_{n=1}^{\infty}\frac{1}{n} & = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+ \frac{1}{2^{k-1}+1}+...+\frac{1}{2^k}+... \\ & = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+ \left(\frac{1}{2^{k-1}+1}+...+\frac{1}{2^k}\right)+...\\ & \geq 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+ \left(\frac{1}{2^{k}}+...+\frac{1}{2^k}\right)+... \\& = 1+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}+... \\& = 1+\sum_{n=0}^{\infty}\frac{1}{2} \end{split} \end{equation*}

and you can do this grouping infinitely many times which gives you infinitely many $\left(\frac{1}{2}\right)$s in the sum, which implies our lower bound diverges which implies our original sum diverges.

Answer 2

"then so too should its sum converge"

If you believe this (and it's reasonable [somewhat] although incorrect assumption) the that would mean the sum converges, not diverges. But no, $a_n\to 0$ does not mean $\sum a_n$ converges and $\sum \frac 1n$ is a classic counterexample. Reread the argument it should make sense.

Thing is we can group the terms $\frac 11, \frac 12, \frac 13,.....$ into groups.

$(\frac 11)$ by itself.

$(\frac 12, \frac 13)$ a group of $2$

$(\frac 14, \frac 15, \frac 16,\frac 17)$ a group of $4$.

Each group is twice as big as the group before.

$(\frac 18,...., \frac 1{15})$ as a group of $8$ and

$(\frac 1{16},....., \frac 1{31})$ as a group of $16$..... and so on.

Each group goes from $\frac 1{2^k}.... $ to $\frac 1{2^{k+1} -1}$ and has $2^k$ terms for some integer $k$.

So what is the sum of each group?

$\frac 1{2^k} + \frac 1{2^k + 1} + \frac 1{2^k+ 1} + ..... + \frac 1{2^k + (2^k-1)} = $ what?

Well, I don't know. But notice each term $\frac 1{2^k + i} > \frac 1{2^{k+1}}$ so

$\frac 1{2^k} + \frac 1{2^k + 1} + \frac 1{2^k+ 1} + ..... + \frac 1{2^k + (2^k-1)} > $

$\frac 1{2^{k+1}} +\frac 1{2^{k+1}}+\frac 1{2^k+ 1}....+\frac 1{2^{k+1}} =$

$\underbrace{\frac 1{2^{k+1}} +\frac 1{2^{k+1}}+\frac 1{2^k+ 1}....+\frac 1{2^{k+1}}}_{\text{there are }2^k\text{ of them}}=$

$ 2^k\times \frac 1{2^{k+1}} = \frac 12$.

So the sum of each group is more than $\frac 12$.

So the sum of $\sum_{j=1}^{\infty} \frac 1n =$

$(\frac 11) + $

$(\frac 12 + \frac 13) + $

$(\frac 14 + \frac 15 + \frac 16 + \frac 17) + $

$(\frac 18+ ..... + \frac 1{15}) + $

$(\frac 1{16} + ..... + \frac 1{31}) + $

$......$

$(\frac 1{2^k} + \frac 1{2^k + 1} + ....... + \frac 1{2^{k+1} -1} ) + $

$..... $

which is the sum of all the sums of the groups. But the sums of each of the groups is more than $\frac 12$.

so....

$\sum_{j=1}^{\infty} \frac 1n =$

$(\frac 11) + $

$(\frac 12 + \frac 13) + $

$(\frac 14 + \frac 15 + \frac 16 + \frac 17) + $

$(\frac 18+ ..... + \frac 1{15}) + $

$(\frac 1{16} + ..... + \frac 1{31}) + $

$......$

$> $

$\frac 12 + \frac 12 + \frac 12 + \frac 12 + ..... \frac 12 + .....$

which is the sum of an infinite number of $\frac12$s.

Which clearly diverges DESPITE then fact that $\frac 1n \to 0$.

$\sum_{j=1}^{\infty} \frac 1n =$

$(\frac 11) + $

$(\frac 12 + \frac 13) + $

$(\frac 14 + \frac 15 + \frac 16 + \frac 17) + $

$(\frac 18+ ..... + \frac 1{15}) + $

$(\frac 1{16} + ..... + \frac 1{31}) + $

$......$

$> $

$(\frac 12) + $

$(\frac 14 + \frac 14) + $

$(\frac 18 + \frac 18 + \frac 18 + \frac 18) + $

$(\frac 1{16}+ ..... + \frac 1{16}) + $

$(\frac 1{32} + ..... + \frac 1{32}) + $

$......$

$=$

$1\times \frac 12+$

$2\times \frac 14 +$

$4\times \frac 18 +$

$8\times \frac 1{16}+$

$16\times \frac 1{32}+$

$...$

$2^k*\frac 1{2^{k+1}} +$

$.....$

$=$

$\frac 12 + \frac 12 + \frac 12 + \frac 12 + ..... \frac 12 + .....$

$\to \infty$.

Answer 3

So a different approach is using Cauchy's criteria for series convergence. It states that if for every $\varepsilon>0$ there exists $n_{0}\in\mathbb{N}$ so that for every $p,m>n_{0}$, $|\sum_{n=m}^{p}an|<\varepsilon$, the series converges. Now as was proved above, you can always find enough elements the series so that their sum is greater than $\frac{1}{2}$. So, for $\varepsilon=\frac{1}{2}$ we have that for every $n_{0}$ we can find $p,m>n_{0}$ such that $|\sum_{n=m}^{p}\frac{1}{n}|>\frac{1}{2}$. This contradicts Cauchy's criteria, and therefore the series diverges.

Answer 4

You can show divergence of a series by finding another series with smaller terms but that is known to diverge (test by comparison).

Here the auxiliary series is found by forming groups of terms and replacing every element in the group by the smallest of them, e.g.

$$\frac15+\frac16+\frac17+\frac18\ \to\ \frac18+\frac18+\frac18+\frac18.$$

Hence,

$$1+\frac12+\color{blue}{\frac13+\frac14}+\color{green}{\frac15+\frac16+\frac17+\frac18}+\cdots>1+\frac12+\color{blue}{\frac24}+\color{green}{\frac48}+\cdots$$ and the auxiliary series obviously diverges.

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Addendum:

It is known that the partial sums grow like the logarithm of $n$. For this reason, when you group the terms and double the group size every time, you can expect that the groups have approximately a constant sum.