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?
 A: 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,
$$
\sqrt {\ln 2} \sum\limits_{k = 1}^\infty  {\frac{{\sqrt {k + 1} }}{{2^k }}}  \le \sqrt {3\ln 2}<\sqrt{3\cdot 0.75} = \frac{3}{2}.
$$
A: $$S=\sum_{k=1}^{\infty }\frac{\sqrt{k+1}}{2^k}=\sum_{k=1}^{\infty }\sqrt{k}\frac{\sqrt{1+\frac{1}{k}}}{2^k}$$ So
$$\sqrt{k+1}=\sum_{p=0}^\infty \binom{\frac{1}{2}}{p} \,k^{\frac{1}{2}-p}$$ Swith the summations to face
$$S=\sum_{p=0}^\infty \binom{\frac{1}{2}}{p} \text{Li}_{p-\frac{1}{2}}\left(\frac{1}{2}\right)$$
which converges very fast. Computing a few partail sums
$$T_n=\sum_{p=0}^n \binom{\frac{1}{2}}{p} \text{Li}_{p-\frac{1}{2}}\left(\frac{1}{2}\right)$$
$$\left(
\begin{array}{cc}
n & T_n \\
 1 & 1.75032 \\
 2 & 1.67221 \\
 3 & 1.70690 \\
 4 & 1.68638 \\
 5 & 1.70038 \\
 6 & 1.69000 \\
 7 & 1.69811 \\
 8 & 1.69154 \\
 9 & 1.69701 \\
 10 & 1.69237
\end{array}
\right)$$
Even with $T_1$ you have the inequality.
Edit
If you enjoy special functions
$$S=\frac{1}{2} \Phi \left(\frac{1}{2},-\frac{1}{2},2\right)$$ where appears the Lerch transcendent function. Its numerical value is $1.69451$.
