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Let $\{a_n\}$ be an increasing sequence greater than 1. Prove that: $\sum\limits_{n=1}^\infty\frac{a_{n+1}-a_n}{a_n\ln a_{n+1}}$ converges $\iff \{a_n\}$ is bounded.

$(\Longleftarrow)$ $\sum\limits_{n=1}^m\frac{a_{n+1}-a_n}{a_n\ln a_{n+1}}\le\sum\limits_{n=1}^m\frac{a_{n+1}-a_n}{a_1\ln a_1}=\frac{a_{m+1}-a_1}{a_1\ln a_1}$, getting the result.

I'm a little bit stuck on $(\implies)$. I know that if $\{a_n\}$ diverges, then $\sum\limits_{n=1}^\infty\frac{a_{n+1}-a_n}{a_n\ln a_n}$ diverges, using integrals to give an estimate. But the same method seems not to work here.

dezdichado
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Saunders
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  • @THeo Sorry, I don't grab what you mean. I prove that the partial sum is bounded, so the series converges. – Saunders Jan 12 '22 at 02:29
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    No, that was my bad. You're right; I just mistook the $a_1$ for an $a_m$. It's also important, with this approach, to point out that the terms of the sum are non-negative, so that the partial sums are monotone increasing. Without this, even bounded partial sums may not converge. – Theo Bendit Jan 12 '22 at 02:31

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Suppose $a_n$ is not bounded, while your sum is convergent. We can assume $a_1 = 1$ for simplicity and let $a_{n+1} = a_n e^{b_n}$ so that $a_{n+1} = e^{b_1+b_2+\dots b_n}$ with $b_i >0.$ Then, $$\infty > \sum_n\dfrac{a_{n+1}-a_n}{a_n\ln(a_{n+1})} = \sum_n\dfrac{e^{b_n}-1}{b_1+b_2+\dots+b_n}>\sum_n\dfrac{b_n}{b_1+b_2+\dots+b_n}.$$

On the other hand, $a_n\to\infty$ which means $\sum_n b_n = \infty$. However, it's a simple exercise to show that $\sum_n b_n=\infty $ implies the divergence of $$\sum_n\dfrac{b_n}{b_1+b_2+\dots +b_n}.$$ See for example here

dezdichado
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