# $\sum a_n\text{ converges}\implies\lim_{n\to\infty}\left(a_0 a_n+a_1 a_{n-1}+a_{\lfloor\frac{n}{2}\rfloor}a_{\lceil\frac{n}{2}\rceil}\right)=\ ?$

Does $$\ \displaystyle \sum_{n\geq 0} a_n\ \text{ converges}\ \implies\ \lim_{n\to\infty}\left( a_0 a_n + a_1 a_{n-1} + a_2 a_{n-2} + a_{ \big\lfloor \frac{n}{2} \big\rfloor } a_{ \big\lceil \frac{n}{2} \big\rceil } \right)\$$ converges?

And if it doesn't converge for conditionally convergent series $$\ \displaystyle \sum_{n\geq 0} a_n\$$, then does it converge for absolutely convergent series $$\ \displaystyle \sum_{n\geq 0} a_n\$$?

If we try to use Cauchy-Schwarz, we get:

$$\left( a_0 a_n + \ldots + a_{ \big\lfloor \frac{n}{2} \big\rfloor } a_{ \big\lceil \frac{n}{2} \big\rceil } \right)^2$$

$$\leq \left( {a_0}^2 + {a_1}^2 + \ldots + {\left( a_{ \big\lfloor \frac{n}{2} \big\rfloor }\right)}^2 \right) \left( {a_n}^2 + {a_{n-1}}^2 + \ldots + {\left( a_{ \big\lceil \frac{n}{2} \big\rceil }\right)}^2 \right).$$

On the right-hand side, the first bracket could $$\ \to \infty,\$$ and I'm not sure what you can say about the second bracket, but even if you could show that the second bracket $$\ \to 0\$$ as $$\ n\to\infty\$$ (which I am not sure of), the right hand side is still indeterminate, so C-S doesn't seem to help.

Also for example if $$\ a_n = \frac{1}{n^2},\$$ then I'm not sure what $$\ \lim_{n\to\infty}\left( a_0 a_n + a_1 a_{n-1} + a_2 a_{n-2} + a_{ \big\lfloor \frac{n}{2} \big\rfloor } a_{ \big\lceil \frac{n}{2} \big\rceil } \right)\$$ is. Perhaps this can be made into an integral? Although I'm not sure how to do this either.

• If the series is absolutely convergent, then I think your CS argument shows the expression in question does converge. See here. If the sequence is not absolutely convergent, say a sequence with oscillating positive/negative terms, then maybe you can show the expression in question oscillates between something with positive lower bound and something with negative upper bound. Commented Jan 17, 2022 at 12:49
• @RodLaver Yes and the limit will be zero, because $$\left\lvert\sum_{k=0}^{\lfloor\frac n2\rfloor}a_k\,a_{n-k}\right\rvert\le\left(\sup_{\lceil\frac n2\rceil\le k\le n}|a_k|\right)\sum_{k=0}^{\lfloor\frac n2\rfloor}|a_k|,$$ where the $\sup$ tends to $0$ and the $\sum$ is bounded. I suspect one can build a counterexample when the series is not absolutely convergent. Commented Jan 17, 2022 at 12:53
• This is strongly related to the Cauchy product of series and Merten's theorem. If $\sum a_n$ converges absolutely then the Cauchy product $\sum c_n$ of $\sum a_n$ with itself is convergent, which implies that $c_n \to 0$. Commented Jan 17, 2022 at 14:03

I believe it does converge for absolutely convergent series, but not necessarily for simply convergent ones.

For non-absolute convergent series a counterexample is:

$$a_n = \frac{(-1)^n}{\sqrt{n}}$$

For absolutely convergent $$a_n$$ introduce $$b_n$$:

$$b_n = \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor} a_j a_{n-j}$$

$$b_{2m} = \sum_{j=0}^m a_j a_{m-j} = \frac{1}{2}\sum_{j=0}^{2m} a_j a_{m-j} + \frac{1}{2} a_m^2$$ $$b_{2m+1} = \sum_{j=0}^m a_j a_{m-j} = \frac{1}{2}\sum_{j=0}^{2m+1} a_j a_{m-j}$$

Using $$f_n = \sum_{j=0}^n a_j a_{n-j}$$ we get: $$\left|b_n - f_n\right| \leq \left|a^2_{\left\lfloor\frac{n}{2}\right\rfloor}\right|$$ which means that $$b_n$$ converges if and only if $$f_n$$ converges. But $$f_n$$ converges because $$2f_n = \left(\sum_{j=0}^n a_j \right)^2 - \sum_{j=0}^n a_j^2$$ which proves the convergence of $$b_n$$ as well.

• +1. The comment to the Q by @nejimban also shows that we do have convergence to $0$ if $\sum a_n$ is absolutely convergent. Also I think we may have convergence to $0$ for some conditionally convergent $\sum a_n$ . Perhaps if $a_n=(-1)^n/(n+1)?$ Commented Jan 17, 2022 at 13:45
• @DanielWainfleet Indeed, in this case $$\sum_{k=0}^{\lfloor\frac n2\rfloor}a_k\,a_{n-k}=\frac{(-1)^n}{n+2}\sum_{k=0}^{\lfloor\frac n2\rfloor}\left(\frac1{k+1}+\frac1{n-k+1}\right)\sim\frac{(-1)^n}n\log n.$$ Commented Jan 17, 2022 at 14:44
• (In the $a_n:=\frac{(-1)^n}{\sqrt{n+1}}$ case we can show that $\sum\limits_{k=0}^{\lfloor\frac n2\rfloor}a_k\,a_{n-k}\sim\frac{(-1)^n\pi}2$ using a Riemann sum argument.) Commented Jan 17, 2022 at 14:51
• I think there is a mistake/typo in the answer. If $$b_n = \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor} a_j a_{n-j}$$ then $$b_{2m} \neq \sum_{j=0}^m a_j a_{m-j}.$$ In the first sum, $b_6 = a_0 a_6 + a_1 a_5 + a_2 a_4 + a_3^2,\$ whereas in the second sum, $b_6 = a_0 a_3 + a_1 a_2 + a_2 a_1 + a_3 a_0.$ Commented Feb 4, 2022 at 11:50
• Thahk you @AdamRubinson. You are right, there was a typo there, which I've just fixed. Commented Feb 4, 2022 at 20:18