Can we prove the divergence of $S=\frac{1}2+\frac{1}3+\frac{1}5+\frac{1}7+\frac{1}{11}\cdots$ like this?

Question

We know that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$ diverges.

So, $T=\frac{1}{n.2}+\frac{1}{n.3}+\frac{1}{n.4}+\frac{1}{n.5}+\cdots$=$\frac{1}{n}$$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots)$ diverges too for any natural number $n$.

Now, we can easily show that $S>T$ since $n$ can take greater value than the difference between the denominators of any two consecutive terms of $S$.

Therefore, $S$ diverges too.

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EDIT: $S$ is the sum of the reciprocals of the primes.

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EDIT 2: $n$ can take any natural numbered value greater than the largest distance between any two consecutive primes.

Answer 1

Your proof is not correct. You say:

$n$ can take greater value than the difference between the denominators of any two consecutive terms of $S$

It is true that if $p_1,p_2$ are two consecutive terms of $S$ (i.e., two consecutive primes) then we can find some $n$ such that $n>p_2-p_1$. However, in order to use this to show that $S$ diverges, we would need to find an $n$ such that $n>q_2-q_1$ for all pairs of consecutive primes $q_1,q_2$.

This is impossible. Indeed, for any $n$, the numbers $n!+2,n!+3,\dots,n!+n$ are not prime and so there must be a pair of consecutive primes that differ by at least $n$.

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If your claim were true, then the sequence $S$ would diverge at least as quickly as any sequence $T_n$. But we know this to be untrue: the harmonic series diverges like $\log(n)$, while the sum of the reciprocals of the primes diverges like $\log(\log(n))$.

Answer 2

Let $$S_n=\sum_{k=1}^n \frac {1}{k} $$ then

$$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac 1k $$

$$>n.\frac {1}{2n}=\frac 12$$

thus $(S_n) $ is not Cauchy and diverges