What is $x$?

It seems it's not an $ x. $

Obviously, CatalinBordea wrote: Let be two nonpositive real numbers $ a,b $ ... should be read as: Quote: Let be two nonnegative real numbers $ a,b $ ... Denote $s_n:=x_1+x_2+\dots+x_n$. Then it holds $|s_n|\leq Cn^a,\forall n\in \mathbb{N}$, where $C$ is some positive constant. We want to estimate $$D_{m,n}:=\left| \frac{x_m}{m^b}+\frac{x_{m+1}}{(m+1)^b}+\dots+\frac{x_n}{n^b}\right|, m<n$$Putting in the above expression $x_k=s_{k}-s_{k-1}, k=m,m+1,\dots,n$ , we get: $$D_{m,n}= \left|s_m\left(\frac{1}{m^b}-\frac{1}{(m+1)^b}\right)+\dots+s_n\left(\frac{1}{n^b}-\frac{1}{(n+1)^b}\right)+\frac{s_{n+1}}{n^b}-\frac{s_{m-1}}{m^b}\right|$$Using $|s_k|\leq Ck^a$, it yields: $$\left|s_k\left(\frac{1}{k^b}-\frac{1}{(k+1)^b}\right)\right|\leq Ck^a\frac{\left( 1+1/k\right)^b-1}{(k+1)^b}\leq \frac{C(b)}{k^{1+b-a}}$$where $C(b)$ is a constant depending only on $b$. Since $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{1+\varepsilon}}$ is convergent for $\varepsilon>0$, it easily follows $D_{m,n}$ can be made as small as we want for sufficiently large $m,n$. It means the series $\sum_{k=1}^{\infty}\frac{x_k}{k^b}$ is convergent.