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I have a question related to the following proof of the divergence of the harmonic series.

Proof

Towards a contradiction, suppose that $\sum_{n=1}^\infty\frac{1}{n}<\infty$ (denote the value of the sum as $\ell$). Then $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dotsb=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}=\frac{\ell}{2}\tag{1} $$ and thus $$ 1+\frac{1}{3}+\frac{1}{5}+\dotsb=\sum_{n=1}^\infty \frac{1}{2n-1}=\frac{\ell}{2}\tag{2}. $$ The author then writes this is a contradiction since $1>1/2$, $1/3>1/4$, $1/5> 1/6$ and so on. $\blacksquare$

Question

I don't see how the chain of inequalities mentioned above implies a contradiction. If we let $s_m=\sum_{n=1}^m\frac{1}{2n}$ and $t_m=\sum_{m=1}^n\frac{1}{2n-1}$ be the sequences of partial sums then we have that $$ s_m<t_m \quad (m\geq 1)\implies\lim_{m\to\infty} s_m\le \lim_{m\to \infty} t_m\tag{3} $$ If the right-hand inequality in (3) was strict we would have a contradiction. But in general we cannot conclude the inequality is strict (for example seee $1-n^{-1}<1$ for all $n\geq 1$). Hence there must be something that I am missing.

Any help is appreciated.

José Carlos Santos
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Sri-Amirthan Theivendran
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  • Alternative one line proof:Let $a_n=\sum_{i=1}^{n}\frac{1}{i}$ and $$\forall n\quad\left|a_{2n}-a_{n}\right|=\sum_{i=n+1}^{2n}\frac{1}{i}>\sum_{i=n+1}^{2n}\frac{1}{2n}=\frac{1}{2}$$hence $a_n$ divergent – ℋolo Aug 06 '18 at 22:57

3 Answers3

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What the author uses is this: if we have two convergent series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ and if $(\forall n\in\mathbb N):a_n>b_n$, then $\sum_{n=1}^\infty a_n>\sum_{n=1}^\infty b_n$. This is indeed true, because$$\left(\sum_{n=1}^\infty a_n\right)-\left(\sum_{n=1}^\infty b_n\right)=\sum_{n=1}^\infty(a_n-b_n)>0,$$since it is the sum of a series of numbers greater than $0$.

José Carlos Santos
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You are correct that $s_m< t_m\ \forall m\Rightarrow\lim s_m\le \lim t_m$ does not by itself yield a contradiction. However, it is easy to prove the stronger statement: $$ s_m+\frac12\le t_m\qquad\text{for all $m$}, $$ which does yield a contradiction.

grand_chat
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  • That certainly works! I won't downvote it, but it doesn't really help the OP to understand the OP's reasoning any better. – Cameron Buie Aug 06 '18 at 23:01
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On one hand, the sum of odd terms equals half the harmonic series because the sum of even terms equals $\ell/2$. But on the other hand, the chain of inequalities imply that every odd term is actually greater than their even counterpart, so the sum of odd terms is also greater than the sum of even terms, which equals $\ell/2$. Hence a contradiction.

Rócherz
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