# Is there any other way to prove this fact? (non-existence of slowest diverging series)

Let $a_n>0$ and $S_n=\sum_{k=1}^{n}a_n$. If $\lim_{n \rightarrow \infty}S_n = +\infty$, then $\sum_{n=1}^{\infty}\frac{a_n}{S_n}=+\infty$. I think this is an important example because it tells us that there exists no series which diverge slowest. So I want to verify this fact in different aspect.

I know a method which uses Cauchy's Theorem. For any $n \in \bf N$ , choose a sufficiently large $p \in \bf{N}$. we have $$\sum_{k=n+1}^{n+p}\frac{a_k}{S_k}\geq \frac{S_{n+p}-S_{n}}{S_{n+p}}\geq \frac{1}{2}$$

Is there any other approach to it? Thanks very much.

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That's probably the simplist way. See this also (I assume this is the argument you're using). – David Mitra Jul 15 '13 at 12:27
If you just want to prove that there is no slowest diverging series, you can see that you can actually choose the increasing sequence $S_n$, and define $a_n=S_n-S_{n-1}$. Then, for any sequence $S_n$ you can get the slower diverging sequence $\sqrt{S_n}$. – Rodrigo Apr 30 '14 at 4:57
Is that answer your question? – Edwin Sep 11 '14 at 3:25

First : $a_n= S_n - S_{n-1}$

Let $U_n$ be : $U_n = \frac{S_n- S_{n-1}}{S_n} = 1 - \frac{S_{n-1}}{S_n}$

Then : $\ln(S_n) -\ln(S_{n-1}) = -\ln(1-U_n)$

Suppose the series of general term $(U_n)$ converges:

$\implies (U_n) \rightarrow 0$

$\implies$ $\ln(S_n) - \ln(S_{n-1}) = U_n + o(U_n)$

It means that $( \ln(S_n) - \ln(S_{n-1}) )$ is the term of a convergent series, thus implying that $(S_n)$ is a converging sequence, which is absurd.

Hence the series of general term $(U_n)$ diverges, always.

Now one interesting question is: what c does make $U_n(c) = \frac{a_n}{S_n^c}$ converge? You know that $c \geq 1$ , but is it the inf?

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