Does the series $\frac{1}{2}\ln 2+\frac{1}{4}\ln 3+\frac{1}{8}\ln 4+\frac{1}{16}\ln 5+\cdots$ converge? It seems to me, by trial and error, that this series converges to $1$. Is this true? How do I prove it either way?
 A: Note 
$$
\sum_{k=2}^\infty\frac{\ln(k+1)}{2^k}\leq \sum_{k=2}^\infty\frac{k}{2^k}=3/2
$$
where the final equality is the usual geometric series game. Thus your series converges, since omitting finitely (here 1) many terms has no effect on convergence.
A: You can rewrite as
$$\sum_{n=2}^{+\infty} \frac{\log n}{2^{n-1}}$$
which clearly converges e.g. for ratio comparison with $\sum\frac{1}{n^2}$.
A: It is obviously convergent but it does not converge to $1$. By Frullani's integral we have
$$S=\sum_{n\geq 1}\frac{\log(n+1)}{2^n}=\int_{0}^{+\infty}\sum_{n\geq 1}\frac{e^{-x}-e^{-(n+1)x}}{2^n}\cdot\frac{dx}{x}=\int_{0}^{+\infty}\frac{2(1-e^{-x})}{(2e^x-1)x}\,dx$$
on the other hand
$$ 2S = \sum_{n\geq 1}\frac{\log(n+1)}{2^{n-1}}=\log(2)+\sum_{n\geq 1}\frac{\log(n+2)}{2^n}   $$
$$ S = \log(2)+\sum_{n\geq 1}\frac{1}{2^n}\log\left(1+\frac{1}{n+1}\right)$$
$$2S=\log(6)+\sum_{n\geq 1}\frac{1}{2^n}\log\left(1+\frac{1}{n+2}\right) $$
$$ S = \log(3)+\sum_{n\geq 1}\frac{1}{2^n}\log\left(1-\frac{1}{(n+2)^2}\right)$$
$$\begin{eqnarray*} S &\geq& \log(3)+9\sum_{n\geq 1}\frac{\log(8)-\log(9)}{2^n (n+2)^2}\\&=&36\log^2(2)\log(3)-54\log^3(2)+\left(9\pi^2-\frac{243}{4}\right)\log(2)+\left(\frac{83}{2}-6\pi^2\right)\log 3\\ &> & 1.0149.\end{eqnarray*}$$
