Suppose for some $\epsilon>0$ that the partial sums of a series are $\sum_{j=1}^na_j<n^{1-\epsilon}$. 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 $\displaystyle a_n<\sum_{j=0}^na_j<n^{1-\epsilon}$, 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?

  • $\begingroup$ Are all the $a_n\ge0$? Is $a_n\ge a_{n+1}$? $\endgroup$
    – robjohn
    Apr 4, 2020 at 3:22
  • $\begingroup$ @robjohn no such details were given unless they are somehow subtly implied by the partial sum condition $\endgroup$ Apr 4, 2020 at 3:29
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 4, 2020 at 3:31

2 Answers 2


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 $\epsilon\gt0$.

  • $\begingroup$ Aha! It was the very last step that I was not seeing $\endgroup$ Apr 4, 2020 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


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