Generating Function question, need help Let $$h_{n} = \sum_{k=0}^{n} \dbinom{n}{k} \frac{2^{k+1}}{k+1}.$$
If $$S= \sum_{n=0}^{\infty} \frac{h_{n}}{n!},$$
find $\lfloor S \rfloor $.
I calculated the first few values of n. I got,
when n=0, we get 2
when n=1, we get 4
when n=2, we get 26/3
i dont see a pattern
 A: Hint. Notice that $$\sum_{k=0}^n \binom{n}{k}\frac{2^{k+1}}{k+1}=\frac{1}{n+1}\cdot \left(3^{n+1}-1\right)$$ Eventhough some combinatorial argument might work in order to prove it, I think that the easiest is considering $\int_0^1 x^k dx = \frac{1}{k+1}$.
In order to finish the problem, simply recall that $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ I obtained $e^3-e\approx 17.37$
A: Hint to start:
$$
\eqalign{
  & \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)} {\left( \matrix{
  n \hfill \cr 
  k \hfill \cr}  \right)x^{\,k} }  = \left( {1 + x} \right)^{\,n}   \cr 
  & \int_{x = 0}^t {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)} {\left( \matrix{
  n \hfill \cr 
  k \hfill \cr}  \right)x^{\,k} dx} }  = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)}
 {\left( \matrix{
  n \hfill \cr 
  k \hfill \cr}  \right){{t^{\,k + 1} } \over {k + 1}}}  =   \cr 
  &  = \int_{x = 0}^t {\left( {1 + x} \right)^{\,n} dx}
  = {1 \over {n + 1}}\left( {\left( {1 + t} \right)^{\,n + 1}  - 1} \right) \cr} 
$$
