$u_n$ converges if and only if $\frac1{\sum_{k=0}^n a_k} \sum_{k=0}^n a_ku_k $ converges. 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 $\liminf_{n\rightarrow \infty} \frac{a_n}{ \sum_{k=0}^n a_k} >0.$
 A: **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$$\frac1{s_n} \sum_{k=0}^n a_k u_k = \frac1{s_n} \sum_{j=0}^{j_n} a_{\chi(j)}\\ \leqslant \frac1{s_{\chi(j_n)}} \left( \sum_{k=0}^{\chi(j_n-1)} a_k \right) + \frac{a_{\chi(j_n)}}{s_{\chi(j_n)}} \underset{n\to\infty}\longrightarrow 0+0.$$
