0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Rod Laver
    Jan 17, 2022 at 12:49
  • 2
    $\begingroup$ @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. $\endgroup$
    – nejimban
    Jan 17, 2022 at 12:53
  • 1
    $\begingroup$ 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 $c_n \to 0$. $\endgroup$
    – Martin R
    Jan 17, 2022 at 14:03

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ +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 $a_n=(-1)^n/(n+1)?$ $\endgroup$
    – DanielWainfleet
    Jan 17, 2022 at 13:45
  • $\begingroup$ @DanielWainfleet Indeed, in this case $$\sum_{k=0}^{\lfloor\frac n2\rfloor}a_k\,a_{n-k}=\frac{(-1)^n}{n+2}\sum_{k=0}^{\lfloor\frac n2\rfloor}\left(\frac1{k+1}+\frac1{n-k+1}\right)\sim\frac{(-1)^n}n\log n.$$ $\endgroup$
    – nejimban
    Jan 17, 2022 at 14:44
  • 1
    $\begingroup$ (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.) $\endgroup$
    – nejimban
    Jan 17, 2022 at 14:51
  • 1
    $\begingroup$ 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, $b_6 = a_0 a_3 + a_1 a_2 + a_2 a_1 + a_3 a_0.$ $\endgroup$
    – Adam Rubinson
    Feb 4, 2022 at 11:50
  • $\begingroup$ Thahk you @AdamRubinson. You are right, there was a typo there, which I've just fixed. $\endgroup$
    – blamocur
    Feb 4, 2022 at 20:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .