How to find the value of $\sum_{n=1}^{\infty} \frac{\ln(1+n)}{2^n}.$ Recently I'm studying about sequences and series(real analysis) and got stuck in the above problem. I tested for the convergence of the given series: $\sum u_n, $ where $u_n=\frac{\ln(1+n)}{2^n}$ and $u_{n+1}=\frac{\ln(2+n)}{2^{n+1}}$ . Thus, $$\lim_{n\to\infty} \frac{u_{n+1}}{u_{n}}$$
$$=\frac12\lim_{n\to\infty}\frac{\ln(2+n)}{\ln(1+n)}\;\left(\frac{\infty}{\infty}\text{ form}\right)$$
$$=\frac12\lim_{n\to\infty}\frac{1+n}{2+n},\text{ using L'Hospital's rule}$$
$$=\frac12\cdot1$$
$$=\frac12<1.$$
So by D'Alembert's ratio test, $\sum u_n$ converges.
But can't think of a way to evaluate the given infinite sum due to my little knowledge. Please suggest something regarding this.. Thanks in advance. 
 A: Using the integral representation of the logarithm 
$$
\ln x = \int\limits_0^\infty  {{{e^{\, - \,t}  - e^{\, - \,x\,t} } \over t}dt} 
$$
we get
$$
\eqalign{
  & S = \sum\limits_{k = 0}^\infty  {{{\ln \left( {1 + k} \right)} \over {2^{\,\,k} }}}
  = \sum\limits_{k = 0}^\infty  {\int\limits_0^\infty  {{{e^{\, - \,t}  - e^{\, - \,\,t} e^{\, - \,k\,t} } \over {t\,2^{\,\,k} }}dt} }  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {\sum\limits_{k = 0}^\infty  {{1 \over {2^{\,\,k} }}}  - \sum\limits_{k = 0}^\infty  {{{e^{\, - \,k\,t} } \over {2^{\,\,k} }}} } \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {2 - \sum\limits_{k = 0}^\infty  {e^{\, - \,k\left( {\,t + \ln 2} \right)} } } \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {2 - {1 \over {1 - e^{\, - \,\left( {\,t + \ln 2} \right)} }}} \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {2 - {{e^{\,\,\left( {\,t + \ln 2} \right)} } \over {e^{\,\,\left( {\,t + \ln 2} \right)}  - 1}}} \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {{{e^{\,\,\left( {\,t + \ln 2} \right)}  - 2} \over {e^{\,\,\left( {\,t + \ln 2} \right)}  - 1}}} \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{e^{\, - \,t} } \over {t\,}}\left( {{{e^{\,\,t}  - 1} \over {e^{\,\,t}  - 1/2}}} \right)dt}  =   \cr 
  &  = \int\limits_0^\infty  {{{1 - e^{\, - \,t} } \over {t\left( {e^{\,\,t}  - 1/2} \right)}}dt}  \cr} 
$$
So this could be a first step to find some bounds on the value
of $S$ if not a closed form.
