If $\sum_{n=0}^\infty|x_n|^2 $ converges then $\sum_{n=0}^\infty \frac{x_n}{\sqrt{n+1}}$ converges?

Question

Does $(x_n) \in \ell^2$ imply that $\sum_{n=0}^\infty \frac{x_n}{\sqrt{n+1}}$ converges?

I have tried finding an upper bound to the series but have failed so far. I have tried Cauchy-Schwarz, some other futile trials and I tried to find counter examples by hand, but I cannot prove it. I have a strong feeling it is true because it even holds for $x_n=1/n$.

Any hints welcome!

Answer 1

This is not true. Let $x_n=\frac 1 {\sqrt n \ln \, n}$ for $n >1$. Then $\{x_n\} \in \ell^{2}$ but $\sum \frac {x_n} {\sqrt {n+1} }=\infty$. [Use the fact that $\sum \frac 1 {n (\ln\, n)^{2}} <\infty$ but $\sum \frac 1 {n (\ln\, n)}=\infty$].