How to show that $\sqrt{\ln 2}\times\sum_{k=1}^{\infty}\frac{\sqrt{k+1}}{2^k}<\frac{3}{2}$? I know that $\sum_{k=1}^{\infty}\frac{\sqrt{k+1}}{2^k}$ is a convergent sequence. Indeed, I can show that $\sum_{k=1}^{\infty}\frac{\sqrt{k+1}}{2^k}\leq \sum_{k=1}^{\infty}\frac{k+1}{2^k}=3$. However, when I multiply $\sum_{k=1}^{\infty}\frac{\sqrt{k+1}}{2^k}$ by $\sqrt{\ln 2}$, I can use Mathematica to show that $\sqrt{\ln 2}\sum_{k=1}^{\infty}\frac{\sqrt{k+1}}{2^k}<\frac{3}{2}$. However, I would like to seek for a mathematical proof. Unfortunately, I cannot prove the assertion rigorously. Could anyone help me?

By the Cauchy–Schwarz inequality
$$
\sum\limits_{k = 1}^\infty  {\frac{{\sqrt {k + 1} }}{{2^k }}}  = \sum\limits_{k = 1}^\infty  {\frac{1}{{2^{k/2} }}\frac{{\sqrt {k + 1} }}{{2^{k/2} }}}  \le \sqrt {\sum\limits_{k = 1}^\infty  {\frac{1}{{2^k }}} } \sqrt {\sum\limits_{k = 1}^\infty  {\frac{{k + 1}}{{2^k }}} }  = \sqrt 3 .
$$
Thus,
$$
\sqrt {\ln 2} \sum\limits_{k = 1}^\infty  {\frac{{\sqrt {k + 1} }}{{2^k }}}  \le \sqrt {3\ln 2}<\sqrt{3\cdot 0.75} = \frac{3}{2}.
$$
