Bound on the infinite sum of logarithms Is it possible to show that
$X=\frac 12 \log 3 + \frac 14 \log 4 + \frac 18 \log 5 + \frac{1}{16} \log 6 + \dots < \log 4$?
I think we can do 
$\frac 12 X= \frac 14 \log 3 + \frac 18 \log 4 + \frac {1}{16} \log 5 + \frac{1}{32} \log 6 + \dots $ 
and subtract the two to get
$\frac 12 X=\frac 12 \log 3 + \frac 14 \log \frac 43 + \frac 18 \log \frac 54 + \frac{1}{16} \log \frac 65 + \dots $
So $X =  \log 3 + \frac 12 \log \frac 43 + \frac 14 \log \frac 54 + \frac{1}{8} \log \frac 65 + \dots < \log 3 + \frac 12 \log \frac 43 + \frac 12 \log \frac 54 + \frac{1}{2} \log \frac 65 + \dots$.
But the right most expression telescopes to $\frac 12 \log 3$ which is absurd! Where is my mistake and how to prove the original statement?
 A: A rigorous approach: let $\displaystyle S_n=\sum_{k=0}^n\frac{1}{2^k}=2-\frac{1}{2^n}$
Performing an Abel transformation yields: 
$$\begin{align}\sum_{k=3}^n \frac{\ln k}{2^k} &= \ln n S_n - \ln 3 S_2 + \sum_{k=3}^{n-1}\ln\left( \frac{k}{k+1}\right)S_k \\ &=2\ln n -\frac{\ln n}{2^n} + \sum_{k=3}^{n-1}\frac{1}{2^k}\ln\left(\frac{k+1}{k}\right) -2\ln n +2\ln3 - \frac 74 \ln 3 \\ &= (2-\frac 74) \ln 3+ \sum_{k=3}^{n-1}\frac{1}{2^k}\ln\left(1+\frac 1k\right)-\frac{\ln n}{2^n} \\ &\le (2-\frac 74) \ln 3+ \sum_{k=3}^{\infty}\frac{1}{k2^k} \\ &\leq  \frac{\ln 3}{4} + \ln 2 - \frac{5}{8} 
\end{align}$$ 
Hence $$\sum_{k=3}^n \frac{\ln k}{2^{k-2}} \leq 4\left( \frac{\ln 3}{4} + \ln 2 - \frac{5}{8} \right) = \ln 48 - \frac 52 <\ln 4$$
I hope you bear no grudge if I don't prove $\displaystyle \ln 48 - \frac 52 < \ln 4$.
Checking with a pocket calculator, $\ln 48 - \frac 52 \sim 1.371$ while $\ln 4\sim 1.386$

You can refine the bound ad libitum if you use sharper inequalities for $\ln(1+x)$.
Using $\ln(1+x)\leq x-\frac{x^2}2+\frac{x^3}3$ and computing $\sum _{k=3}^{\infty } \frac{1}{2^k}\left(\frac{1}{3 k^3}-\frac{1}{2 k^2}+\frac{1}{k}\right)$. with Mathematica yields the much tighter bound:
$$\sum_{k=3}^n \frac{\ln k}{2^{k-2}}\leq \frac{1}{144} \left(42 \zeta (3)-6 \pi ^2-75+8 \log ^3(2)+36 \log ^2(2)-4 \pi ^2 \log (2)+144 \log (2)\right)+\log (3)$$
with $\frac{1}{144} \left(42 \zeta (3)-6 \pi ^2-75+8 \log ^3(2)+36 \log ^2(2)-4 \pi ^2 \log (2)+144 \log (2)\right)+\log (3)\sim \color{red}{1.3397}$
A: Wolfram Alpha gives:
$$\int_1^\infty \log(x+2)/2^x dx = (4 \text{Ei}(-\log(2) x-\log(4))-2^{-x}\frac{\log(x+2))}{(\log(2))} \Big|_1^{\infty}$$
$$=(\log(3)-8 Ei(-\log(8)))/(\log(4)) = 1.04542...$$
where  $$\operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt.\,$$ is the exponential integral.
So by the integral test, you have your bound.
Edit: As someone pointed out in the comments, your right-hand-side series actually does not converge. The reason is that the partial sums all telescope to the form
$$\frac{1}{2}\log 3 + \frac{1}{2}\log n$$
which increases to infinity as n increases. 
