# Suppose that $\sum_{j=1}^na_j<n^{1-\epsilon}$ for $\epsilon>0$. Show that $\sum_{n=1}^\infty \frac{a_n}{n}$ converges.

Suppose for some $$\epsilon>0$$ that the partial sums of a series are $$\sum_{j=1}^na_j. I want to show that $$\sum_{n=1}^\infty\frac{a_n}{n}$$ converges.

I am not sure I know a test to apply to figure out the convergence of this series. Abel's test does not work because $$n^{1-\epsilon}$$ is not a uniform bound. It's not obvious how I would use Kummer's test because I don't know anything about the ratio of successive terms. The only thing I can think is that surely $$\displaystyle a_n<\sum_{j=0}^na_j, so then $$\displaystyle\sum_{n=1}^\infty\frac{a_n}{n}<\sum_{n=1}^\infty\frac{n^{1-\epsilon}}{n}$$ but this isn't very helpful since the one on the right diverges :(

The only other thing I tried was using summation by parts. Let $$b_n=\frac{1}{n}$$ then $$\sum_{j=1}^Na_jb_j=b_N\sum_{j=1}^Na_j + \sum_{n=1}^{N-1}\left(\sum_{j=1}^na_j(b_n-b_{n+1})\right)$$ $$\sum_{j=1}^N\frac{a_j}{j}<\frac{N^{1-\epsilon}}{N}+\sum_{n=1}^{N-1}n^{1-\epsilon}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ <\frac{1}{N^\epsilon}+\sum_{n=1}^{N-1}\left(1-\frac{n}{n+1}\right)$$ But then again, the series on the right is divergent when we let $$N\rightarrow \infty$$ so this was not helpful. Is there a clear test to use? Have I made a critical error?

• Are all the $a_n\ge0$? Is $a_n\ge a_{n+1}$? – robjohn Apr 4 '20 at 3:22
• @robjohn no such details were given unless they are somehow subtly implied by the partial sum condition – Wyatt Kuehster Apr 4 '20 at 3:29
• @WyattKuehster If you allow $a_n \lt 0$, then it's quite simple to come up with a counter-example, e.g., $a_n = -n$, so I suspect at least that is assumed. – John Omielan Apr 4 '20 at 3:31

Suppose $$\sum_{k=1}^na_k\le n^{1-\epsilon}\tag1$$ where $$a_n\ge0$$. Then \begin{align} \sum_{n=1}^\infty\frac{a_n}n &=\sum_{n=1}^\infty\sum_{k=n}^\infty\left(\frac1k-\frac1{k+1}\right)a_n\tag2\\ &=\sum_{k=1}^\infty\sum_{n=1}^k\frac1{k(k+1)}\,a_n\tag3\\ &\le\sum_{k=1}^\infty\frac1{k^2}\,k^{1-\epsilon}\tag4\\ &=\sum_{k=1}^\infty k^{-1-\epsilon}\tag5 \end{align} Explanation:
$$(2)$$: write $$\frac1n$$ as the sum of a telescoping series
$$(3)$$: change the order of summation and $$\frac1k-\frac1{k+1}=\frac1{k(k+1)}$$
$$(4)$$: apply $$(1)$$ and $$\frac1{k(k+1)}\lt\frac1{k^2}$$
$$(5)$$: simplify

The sum in $$(5)$$ converges for $$\epsilon\gt0$$.

• Aha! It was the very last step that I was not seeing – Wyatt Kuehster Apr 4 '20 at 3:36

Denote $$s_n=\sum_{k=1}^n a_j$$ By assumption we have ($$\epsilon>0$$): $$s_n\leq n^{1-\epsilon} \quad (1)$$

Using summation by parts with $$a_n=s_n-s_{n-1}$$ and $$\tfrac{1}{n}-\tfrac{1}{n+1}=\tfrac{n}{n+1}$$ we get (the upper boundary term can be neglected since $$s_n/n$$ goes to zero for large $$n$$ by assumption (1)).

$$\sum_{n=1}^{\infty}\frac{a_n}{n}=\sum_{n=2}^{\infty}\frac{s_n}{n(n+1)}+a_1$$ or by (1) $$\sum_{n=1}^{\infty}\frac{a_n}{n}\leq \sum_{n=2}^{\infty}\frac{1}{n^{\epsilon}(n+1)}+a_1<\infty$$

which means that the series converges