$u_n$ converges if and only if $\frac1{\sum_{k=0}^n a_k} \sum_{k=0}^n a_ku_k $ converges.

Question

Let $(a_n)_{n \in \mathbb{N}} $ be a positive sequence with $a_0\neq0$. Find a necessary and sufficient condition on $(a_n) $ in order that:

for any real sequence $(u_n)_{n \in \mathbb{N}}$, $$\boxed{(u_n)_{n \in \mathbb{N}}\quad \text{converges}\quad \iff\quad \left(\frac1{\sum_{k=0}^n a_k} \sum_{k=0}^n a_ku_k \right)_{n \in \mathbb{N}}\quad \text{converges }}$$

Add 1: Thank you very much @SeverinSchraven , I see now that the condition sought is $\liminf_{n\rightarrow \infty} \frac{a_n}{ \sum_{k=0}^n a_k} >0.$

You surely need that $\sum_{k\geq 0} a_k = \infty$. Otherwise, you have a problem with $u_n =(-1)^n$. — Severin Schraven, Jan 16 '22 at 23:54
However, $\sum_{k\geq 0} a_k=\infty$ is not sufficient as for example $a_k=1/(k+1)$ and $u_n=(-1)^n$ shows. — Severin Schraven, Jan 16 '22 at 23:57
We need $\lim_{n\rightarrow \infty}a_n=\infty$. If this was not the case, then we can pick a subsequence $(a_{n_m})$ and a finite number $\ell = \lim_{m\rightarrow \infty} a_{n_m}=\ell$. Still we have $\sum_{k\geq 0} a_k =\infty$, then pick $u_{n_m}=1$ and zero otherwise and thus $\vert \sum_{k=0}^n a_k u_k \vert \leq C \sum_{k=0}^n a_k$, where $C=\sup a_{n_m} <\infty$. — Severin Schraven, Jan 17 '22 at 00:07
Note that under the condition $\sum_{k\geq 0} a_k = \infty$ the implication $"\Rightarrow"$ holds true by the same proof as for Cesaro's theorem. — Severin Schraven, Jan 17 '22 at 00:22
We need some growth condition on $a_k$. Presumably something along the lines $$ \liminf_{n\rightarrow \infty} \frac{a_n}{ \sum_{k=0}^n a_k} >0. $$ Otherwise we can make a similar construction as in my third comment (lots of zeros and occassional ones in $u$). — Severin Schraven, Jan 17 '22 at 00:36
@SeverinSchraven Would you like to try $a_n = 2^n$ as a specific possibility? — Sarvesh Ravichandran Iyer, Jan 17 '22 at 00:44
@SarveshRavichandranIyer I let it rest a bit for now :) had a lot of fun thinking about it, but I should go back and prepare my class for tomorrow and the grant application also doesn't write itself. If you know how to do it, I would be very curious to see how one cracks this nut. — Severin Schraven, Jan 17 '22 at 00:50
@SeverinSchraven Thanks for your efforts thus far, and good luck for your class. — Sarvesh Ravichandran Iyer, Jan 17 '22 at 00:50
@SarveshRavichandranIyer Thank you :) wil be a lot of fun. I'll be doing Riemann sums and the fundamental theorem of calculus with my students. Also I have great students, it's a joy to teach them. — Severin Schraven, Jan 17 '22 at 01:37
@Jane Did you figure out how to show $"\Leftarrow"$ under the condition with $\liminf$? — Severin Schraven, Jan 17 '22 at 16:58
@Jane Do consider writing a self-answer (with some detail) when you can. — Sarvesh Ravichandran Iyer, Jan 17 '22 at 17:59
It's not my solution but by my French friend, can I give it in French? — Paul, Jan 17 '22 at 19:55
@Jane Probably best if you post it as an answer below, translated into English — Jose Avilez, Jan 17 '22 at 20:11

Paul · Answer 1 · 2022-01-17T21:48:32.347

**It's due to my french friend Calli **

Sufficient character: assumed that $\liminf\limits_{n\to\infty} \frac{a_n}{s_n} >0$.

$\Rightarrow$ : Let $(u_n)$ such that $u_n$ converges. If $(s_n)$ was bounded, so we would have $s_n=O(a_n)=o(1)$, but $s_0=a_0>0$ and $(s_n)$ is increasing, so is absurd. So we have $s_n\to\infty$, et $\frac1{s_n} \sum_{k=0}^n a_k u_k$ converges from generalised Cesàro..

$\Leftarrow$ : Let $(u_n)$ such that $v_n:=\frac1{s_n} \sum_{k=0}^n a_k u_k$ converges to a limit noted $\ell$. Note $p_n := \frac{a_n}{s_n}$. As $\liminf\limits_{n\to\infty} p_n >0$, there exists $c>0$ et $n_0$ such that : $\forall n>n_0,\; p_n \geqslant c$. thus : $\forall n>n_0$, $$\begin{eqnarray*} \frac1{s_n} \frac{s_n-a_n}{s_{n-1}} \sum_{k=0}^{n-1} a_ku_k +\frac{a_n}{s_n} u_n &=& v_n\\[1mm] (1-p_n)v_{n-1} +p_n u_n &=& v_n\\[1mm] (1-p_n)(v_{n-1}-\ell) +p_n (u_n-\ell) &=& v_n-\ell\\[1mm] p_n |u_n-\ell| &\leqslant& |v_n-\ell| + (1-p_n)|v_{n-1}-\ell| \\[1mm] c\, |u_n-\ell| &\leqslant& |v_n-\ell| +|v_{n-1}-\ell| \longrightarrow 0 \end{eqnarray*}$$ thus $u_n\to\ell$.

Necessary character: By Contraposed, suppose $\liminf\limits_{n\to\infty} \frac{a_n}{s_n} =0$, where $s_n :=\sum_{k=0}^n a_k$, and Let's find a counter-example $(u_n)$.

If $(s_n)$ is bounded , then it converges (because it is increasing) and $\sum (-1)^n a_n$ is absolutely convergent. So $\frac1{s_n}\sum_{k=0}^n (-1)^n a_n $ converges while $(u_n) :=((-1)^n)$ diverges. Otherwise, $s_n\to+\infty$. Let $\varphi :\Bbb N\to\Bbb N$ such as $\frac{a_{\varphi(n)}}{s_{\varphi(n)}}\to 0$. We build by recurrence $\psi$ such that $\psi(0)=0$ and, for any $n$, $\psi(j+1)$ is such that $$\frac1{s_{\varphi\circ\psi(j+1)}} \sum_{k=0}^{\varphi\circ\psi(j)} a_k < \frac1{2^j}.$$ and, we put $\chi=\varphi\circ\psi$ and, for any $j$, $u_n :={\bf1}_{\chi(\Bbb N)}(n)$ et $j_n :=\max\{j\in\Bbb N\mid \chi(j) \leqslant n\}$. So $(u_n)$ do not converges, but$$\frac1{s_n} \sum_{k=0}^n a_k u_k = \frac1{s_n} \sum_{j=0}^{j_n} a_{\chi(j)}\\ \leqslant \frac1{s_{\chi(j_n)}} \left( \sum_{k=0}^{\chi(j_n-1)} a_k \right) + \frac{a_{\chi(j_n)}}{s_{\chi(j_n)}} \underset{n\to\infty}\longrightarrow 0+0.$$

$u_n$ converges if and only if $\frac1{\sum_{k=0}^n a_k} \sum_{k=0}^n a_ku_k $ converges.

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