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. 

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, 
$$\sum\limits_{k=1}^\infty \frac{\log(1+k)}{2^k} 
\leq \sum\limits_{k=1}^\infty \left(\frac{1+k}{3 \cdot 2^k}\right) \log 3
=\log 3$$
