# Suppose $\sum \frac{1}{a_n}$ diverges. Then does $\sum \frac{1}{n\Delta a_n}\$ diverge?

Let $$(a_n)$$ be a strictly increasing sequence of positive real numbers, and denote $$\Delta a_n:= a_{n+1} - a_n.$$

Suppose $$\displaystyle\sum \frac{1}{a_n}$$ diverges. Then does $$\displaystyle\sum \frac{1}{n\Delta a_n}\$$ diverge?

My first thought for counter-examples was to try $$a_n = n\log n.$$ But I'm pretty sure $$\displaystyle\sum \frac{1}{n\left( (n+1)\log(n+1) - n\log n \right)}\$$ diverges.

I can't think of a good proof strategy for the affirmative.

• I think it actually converges. You can replace $\log(n+1)$ by $\log n$ and simplify. Commented Sep 3, 2023 at 21:41
• @Dog_69 $\sum \frac{1}{n\left( (n+1)\log(n+1) - n\log n \right)}$ diverges, since $(n+1)\log(n+1) - n\log n\sim\log n.$ Commented Sep 3, 2023 at 21:43
• @AnneBauval yes, sorry, I was thinking about $n^{1+\log n}$ instead. Commented Sep 3, 2023 at 21:46

Lemma 1. If $$a_n>0$$ and $$\sum_{n=1}^\infty \frac1{na_n}<+\infty$$, then $$\sum_{n=1}^\infty\frac1{a_1+a_2+\cdots+a_n}<+\infty.$$
Indeed, if $$(a_n)$$ is a strictly increasing positive sequence and $$\sum_{n=1}^\infty\frac1{n\Delta a_n}<+\infty$$, then we let $$b_n=a_{n}-a_{n-1}>0$$ for $$n\geq2$$ and $$b_1=a_1>0$$, and hence $$\sum_{n=1}^\infty\frac1{nb_n}\leq \frac1{a_1}+\sum_{n=1}^\infty\frac1{n\Delta a_n}<+\infty$$. By Lemma 1, we have $$\sum_{n=1}^\infty\frac1{a_n}=\sum_{n=1}^\infty\frac{1}{b_1+\cdots+b_n}<+\infty$$. This gives an affirmative answer to OP's problem.
Lemma 2. If $$a_n>0$$ and $$\sum_{n=1}^\infty a_n<+\infty$$, then $$\sum_{n=1}^\infty\frac{1}{\sum_{j=1}^n\frac1{ja_j}}<+\infty.$$
By Cauchy-Schwarz, we have $$k=\sum_{j=1}^k1=\sum_{j=1}^k\sqrt{ja_j}\cdot\sqrt{\frac1{ja_j}}\leq \sqrt{\sum_{j=1}^kja_j}\times \sqrt{\sum_{j=1}^k\frac1{ja_j}},$$ then $$\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq \frac1{k^2}\sum_{j=1}^kja_j\leq2\left(\frac1k-\frac1{k+1}\right)\sum_{j=1}^kja_j.$$ Let $$A_k=\frac1k\sum_{j=1}^{k-1}ja_j$$ for $$k\geq2$$ and $$A_1=0$$, then $$\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq2a_k+A_k-A_{k+1}.$$ Summing the above inequality for $$1\leq k\leq n$$ and using $$A_n\geq 0$$ for $$n\geq 1$$, we obtain $$\sum_{k=1}^n\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq2\left(\sum_{k=1}^na_k\right)-A_{n+1}\leq 2\sum_{k=1}^na_k.$$ The proof of Lemma 2 is complete now.