# How to prove that if $\sum _{n=1}^{\infty }a_n\:$ converges then $\sum _{n=1}^{\infty }a_na_{2n}\:$ converges?

How to prove that if $\sum _{n=1}^{\infty }a_n\:$ converges then $\sum _{n=1}^{\infty }a_na_{2n}\:$ converges?

Note: $a_n \in \mathbb R$

I tried to prove it using cauchy criterion The idea was to use the partial sums.

let $\epsilon$>0

There exists $N_1$ such that for any $n>N_1$ and $p\ge1$

$$\left|\sum \:\:\:\:_{k=n+1}^{n+p}a_k\:\right|<\epsilon \:$$ There exists $N_2$ such that for any $n>N_2$ and $p\ge1$

$$\left|\sum \:\:\:\:_{k=n+1}^{n+p}a_{2k}\:\right|<\epsilon \:$$ We will take $N=max\left\{N_1,N_2\right\}$ and then for any $n>N$ and $p\ge1$:

$$\left|\sum \:\:_{k=n+1}^{n+p}a_ka_{2k}\:\right|\le \left|\sum \:\:\:_{k=n+1}^{n+p}\left(max\left\{a_k,a_{2k}\right\}\right)^2\:\right|\le ...<\epsilon$$

But I didn't manage to show the part with the three dots. Maybe it's not the way to prove this. Can I get help please ?

• @zyx can you expand on that?
– user290300
Aug 26, 2016 at 22:12
– zyx
Aug 27, 2016 at 13:06

There might be a problem. Let $\gamma=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ be a non-trivial cube root of $1$. $$a_n= \frac{\gamma^n}{\sqrt{n}}$$ Then $\sum_n a_n$ converges but $$\sum_{n\geq 1} a_n a_{2n} = \sum_{n\geq 1} \frac{\gamma^{3n}}{\sqrt{2}n}= \sum_n \frac{1}{\sqrt{2}n}=+\infty$$

For the real case set $b_n={\rm Re\;} a_n$ then $\sum_n b_n$ converge. However, $${\rm Re} (\gamma^n) {\rm Re} (\gamma^{2n})= ){\rm Re} (\gamma^n) {\rm Re} (\gamma^{-n}) =({\rm Re} (\gamma^n))^2 \geq 1/4$$ so $\sum_n b_n b_{2n}\geq \frac{1}{4\sqrt{2}} \sum_n \frac{1}{n}=+\infty$.

• Note that it's true via Dirichlet test when $a_{2n}$ is decreasing (and real). Aug 26, 2016 at 22:17
• Nicely done! What if the requirement was $a_n \in \mathbb R$? (I have been thinking about a counterexample but failed at finding one)
– user290300
Aug 26, 2016 at 22:17
• @madmatician The statement in my question is true. It was at my final exam yesterday. $a_n \in \mathbb R$ Aug 26, 2016 at 22:19
• For a real example, take $a_n = \sin(2 n \pi/3)/\sqrt{n}$. Aug 26, 2016 at 22:23

As H.H. Rugh's answer points out, the result doesn't hold in general. Here is a proof for the case $a_n>0$:

First note that $a_na_{2n}\leq \frac{a_n^2+a_{2n}^2}{2}$ for all $n$, and $0<a_n^2<a_n$ for all sufficiently large $n$ since $a_n\to 0$, hence $$\sum_{n=1}^{\infty}a_na_{2n}\leq \frac{1}{2}\sum_{n=1}^{\infty}a_n^2+\frac{1}{2}\sum_{n=1}^{\infty}a_{2n}^2\ll \sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}a_{2n}\ll \sum_{n=1}^{\infty}a_n<\infty$$

• No need for $a_n>0$. Just absolute convergence will do
– user290300
Aug 26, 2016 at 22:21
• Sure, that works too. Aug 26, 2016 at 22:21
• @carmichael561 How can the statement be a false statement ? I had it at my exam yesterday. We dont know if $a_n$ is positive for all $n$ Aug 26, 2016 at 22:23
• It is not unheard of for exams to contain mistakes. Aug 26, 2016 at 22:24
• Also note that since $a_n\to 0,$ $|a_na_{2n}| \le |a_n|$ for large $n.$
– zhw.
Aug 26, 2016 at 22:24

Define

$$a_{3k}=0, a_{3k+1}=\frac{1}{\sqrt{3k+1}}, a_{3k+2}=-\frac{1}{\sqrt{3k+2}}$$ then $\sum a_{n}$ converges but not $\sum a_{n}a_{2n}$.