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I was given a proof of the divergence of the harmonic series, but I can't fully understand it (I found other ones that I find exhaustuve, so the problem is not trying to understand why the series diverges)

Here is the proof:

Having: $\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\gt\frac{3}{x}$ for $x\gt1$

Let be $$S=\sum_{n=1}^\infty \frac1n\lt\infty$$

$$S=1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+...$$ $$S = 1+\frac{1}{3-1}+\frac{1}{3}+\frac{1}{3+1}+\frac{1}{6-1}+\frac{1}{6}+\frac{1}{6+1}+...$$

Now, using the previous inequality

$$S\gt1+3(\frac13+\frac16+\frac19+...) =1+1+\frac12+\frac13+...=1+S$$

So we get at the end $S\gt1+S$, as a consequence $S$ can't be finite.

My question is if it is possible to treat an infinite sum like that.

Thanks in advance.

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    $\begingroup$ You can make the proof rigorous by summing a finite number of terms, as a function of $n$. Then if you can show $S_n<S_n+1$, this holds in the limit. But it seems that in fact you don't establish $S_n<S_n+1$ but $S_n<1+S_{3n}$ or something. $\endgroup$
    – user65203
    Jan 20, 2021 at 10:20
  • $\begingroup$ Sorry, i mistyped the greater/less than, I meant $S\gt1+S$ at the end. Now I've adjusted it. $\endgroup$
    – Royolh
    Jan 20, 2021 at 10:27
  • $\begingroup$ Keep the spirit of my answer. $\endgroup$
    – user65203
    Jan 20, 2021 at 10:33
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    $\begingroup$ By working with a finite number of terms I can only establish that $S_n\gt1+S_\frac{n-1}{3}$ So I guess the proof is not valid because you don't get an absurd. $\endgroup$
    – Royolh
    Jan 20, 2021 at 10:44

2 Answers 2

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Though clever, this manipulation is not rigorous. A series converges if the limit of the partial sums converges. If we consider the $7^{\text{th}}$ partial sum, like you did, we do have

$$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17>1+\frac33+\frac36=1+1+\frac12$$

but this is not

$$S_7>1+S_7.$$ Instead, it establishes $$S_7>1+S_2$$ which generalizes to $$S_{3n+1}>1+S_n$$ and this not enough to prove divergence.


The situation is different with the classical proof by groups of doubling sizes,

$$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18>1+\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\frac18 \\=1+\frac12+\frac12+\frac12$$

or

$$S_9>1+\frac{3}2,$$ which generalizes to $$S_{2^n+1}>1+\frac{n}2$$

or

$$S_m>1+\frac{\log_2(m-1)}2.$$

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First let's make some observations. the sequence of partial sums is a sequence of positive integers numbers therefore the sum of the series either diverges or converges. if I prove that the series does not converge then it necessarily diverges.

so let's proceed absurdly. and suppose that the sum of the series converges

so

$$ \exists C\in R$$

$$\forall \epsilon >0\quad \exists n_{1} :\quad \forall n>n_{1}\quad |\sum_{i=1}^{n}\frac{1}{i}-C|<\epsilon$$

if this sum of the series converges$\sum_{i=1}^{\infty}\frac{1}{i}$ then this other sum of the series also converges $\lim_{n\to \infty}\sum_{i=1}^{3n+1}\frac{1}{i}$ because they have the same sequence as the partial sums and furthermore we can say that they converge to the same value. $$ \exists M\in R$$

$$\forall \epsilon >0\quad \exists n_{2} :\quad \forall n>n_{2}\quad |\sum_{i=1}^{3n+1}\frac{1}{i}-M|<\epsilon$$

and so for the uniqueness theorem of the limit $$M=C$$

that prof of the question show

$$\sum_{i=1}^{3n+1}\frac{1}{i}>1+\sum_{i=1}^{n}\frac{1}{i}$$

if now we pass to the limit $$\lim_{n\to\infty}\sum_{i=1}^{3n+1}\frac{1}{i}>1+\lim_{n\to\infty} \sum_{i=1}^{n}\frac{1}{i}$$

but $$M>1+C$$ so $$M\neq C$$

we got the absurd so it is not possible that the sums of the series is convergent, so it is divergent for the previous observation.

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