Prove $\sum_{k=0}^{\infty} \frac {\log(1+k)}{2^k} \le \log3$ Initially I faced with a question $$\sqrt{2\sqrt{3\sqrt{4...\sqrt{N}}}} < 3$$ which is not hard to prove by induction.
But if taking logarithm on both sides of this inequality, we get $\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} < \log3$.
Is there any way to prove this series inequality? Or evaluate this series?
I tried to use series of $\log(1+k)$. And I got $\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} < \sum_{k=0}^{\infty} \frac{k}{2^k} =2$, which gives a larger upper bound. 
A surprising fact is that the more term we use for the series of $\log(1+k)$, the larger upper bound we will get. 
 A: It is easy to show that $n^3 \leq 3^n$ for all $n \in \Bbb{N}$.
Thus $n \leq 3^{\frac{n}{3}}$, i.e., $\log n \leq \frac{n}{3} \log 3$. 
Hence, 
$$\sum\limits_{k=1}^\infty \frac{\log(1+k)}{2^k} 
\leq \sum\limits_{k=1}^\infty \left(\frac{1+k}{3 \cdot 2^k}\right) \log 3
=\log 3$$
A: Hint: Once you have an upper bound, you can try using a couple terms of the actual sequence and then using the bound. So, for any positive integer $N$, 
$$\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} =\sum_{k=0}^N \frac{\log(1+k)}{2^k} + \sum_{k=N+1}^{\infty} \frac{\log(1+k)}{2^k} < \sum_{k=0}^N\frac{\log(1+k)}{2^k} + \sum_{k=N+1}^{\infty} \frac{k}{2^k}$$
A: What about summation by parts? If we let $S=\sum_{k\geq 0}\frac{\log(k+1)}{2^k}$ we have
$$\begin{eqnarray*}\sum_{k=0}^{N}\frac{\log(k+1)}{2^k} &=& (2-2^{-N})\log(N+1)-\sum_{k=0}^{N-1}(2-2^{-k})\log\left(\frac{k+2}{k+1}\right)\\
&=&-2^{-N}\log(N+1)+\sum_{k=0}^{N-1}2^{-k}\log\left(1+\frac{1}{k+1}\right)\end{eqnarray*}$$
hence by letting $N\to +\infty$ we get
$$ \color{red}{S} = 2\sum_{k\geq 1}\frac{\log\left(1+\frac{1}{k}\right)}{2^k}\color{red}{<}\log(2)+2\sum_{k\geq 2}\frac{1}{k 2^k}={3\log(2)-1}\approx \color{red}{1.07944}$$
which is a better bound, equivalent to
$$ \sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots}}}}<\frac{8}{e}. $$
Using $\log(1+x)> \frac{x}{1+\frac{x}{2}}$ on $(0,1)$, we also get the tight lower bound
$$ \color{red}{S >} -\frac{14}{3}+\log(2)+4\sqrt{2}\log(1+\sqrt{2})\approx \color{red}{1.01228}. $$
