<s>[INST] Evaluate $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$ I'm interested in finding a nice closed form expression for the sum $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$. I've tried using the Binomial Theorem to get 
\begin{align*}
\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k & =\int_0^1\frac{(1+x)^n-1}{x} \, dx\\
&=\int_1^2 (1+u+\cdots+u^{n-1}) \, du
\end{align*}
using the substitution $u=1+x$ but I can't quite to simplify this integral either. I have also not been able to come up with a combinatorial approach, which may not exist since the summation and its terms are in general not integers. Any help in evaluating this sum would be appreciated, thanks!
 [/INST] $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k $
This is equal to term by term multiplication of following series:
$∑^n_{k=1}  C^n_k =∑^n_{k=0}  C^n_k-1=(1+1)^n-1=2^n-1$
$∑^n_{k=1}\frac{1}{k}= 1+1/2+1/3 + . . . 1/n =H_n  $
⇒ $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k =(2^n-1)H_n=∑^n_{k=1}\frac{2^k-1}{k}$ 
Note that $(2^n-1)$represent$(∑ ^n_{k=1} C^n_k)$ and $(H_n)$ represent$(∑ ^n_{k=1}\frac{1}{k})$ that is $(∑ ^n_{k=1}\frac{1}{k} C^n_k)=∑ ^n_{k=1}\frac{1}{k}\times∑ ^n_{k=1} C^n_k$(term by term multiplication).For example for n=5:
$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k$  is not$(2^5-1)(1+1/2+1/3+1/4+1/5)=\frac{4247}{60}$, it is calculated as:
$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k=\frac{2^1-1}{1}+\frac{2^2-1}{2}+\frac{2^3-1}{3}+\frac{2^4-1}{4}+\frac{2^5-1}{5}=\frac{887}{60}$
</s>