# Let $a_n$ be a sequence s.t. $\sum a_n$ converges, let $f:N \rightarrow N$ be a function s.t. $f(n) \rightarrow \infty$.Then $\lim f(n) a_{f(n)} = 0$.

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.

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

This is false. Let $$f(n)=n$$ for all $$n$$. Suppose: $$$$a_n = \begin{cases} \frac{1}{n},& \text{n is a square integer}\\ 0, & \text{otherwise} \end{cases}$$$$ Then $$\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$$)].
• If $a_n$ is decreasing then $na_n\to 0,$ as $\sum_{k=[n/2]}^na_k\ge (n/2)a_n.$ In particular $f(n)a_{f(n)}\to 0.$ Jun 10, 2022 at 1:21