I am trying to determine whether the following statement is true or false:

Let $a_n$ be a positive sequence s.t. $\sum_{n=1}^\infty a_n$ converges, and let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function s.t. $f(n) \rightarrow \infty$ as $n\to\infty$.

Then $\lim_{n\to\infty} f(n) a_{f(n)} = 0$.

I believe this statement is true.

It somewhat reminds Cauchy condensation test, so I tried to take inspiration from its proof but it didn't get me far.

I tried to use the squeeze theorem:

$0 \le f(n) a_{f(n)} \le$ ??

But failed to find an upper bound for the above expression.

Any hints will be appreciated.

  • 1
    $\begingroup$ Do you need the $a_n$s to be decreasing? $\endgroup$
    – Mike
    Jun 9, 2022 at 23:45
  • $\begingroup$ @Mike Thats actually the second part of the question $\endgroup$
    – MStudent
    Jun 9, 2022 at 23:46
  • $\begingroup$ What do you mean the second part of the question? $\endgroup$
    – Mike
    Jun 9, 2022 at 23:47
  • $\begingroup$ @Mike There are two questions on my paper, On the second question $a_n$ is decreasing. $\endgroup$
    – MStudent
    Jun 9, 2022 at 23:48

1 Answer 1


This is false. Let $f(n)=n$ for all $n$. Suppose: \begin{equation} a_n = \begin{cases} \frac{1}{n},& \text{$n$ is a square integer}\\ 0, & \text{otherwise} \end{cases} \end{equation} Then $\sum a_n$ converges but $f(n)a_{f(n)}=1$ whenever $n$ is a square.

[If you want $a_n$'s to be strictly positive you can simply add $\frac 1{n^{2}}$ to $a_n$ (for each $n$)].

  • 2
    $\begingroup$ Why was this answer downvoted? $\endgroup$
    – Mike
    Jun 9, 2022 at 23:44
  • 3
    $\begingroup$ Too bad one cannot downvote the downvoters. $\endgroup$ Jun 9, 2022 at 23:45
  • $\begingroup$ Ok thats creative, Thank you. $\endgroup$
    – MStudent
    Jun 9, 2022 at 23:50
  • 1
    $\begingroup$ If $a_n$ is decreasing then $na_n\to 0,$ as $\sum_{k=[n/2]}^na_k\ge (n/2)a_n.$ In particular $f(n)a_{f(n)}\to 0.$ $\endgroup$ Jun 10, 2022 at 1:21
  • $\begingroup$ @RyszardSzwarc Yes, that is a standard result. But OP doesn't want extra assumptions. $\endgroup$ Jun 10, 2022 at 4:57

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