How find this sum $f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}$ Find this sum closed form
$$f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}$$
My idea: since $$\dfrac{1}{i}=\int_{0}^{1}x^{i-1}dx$$
so
$$f(n)=\sum_{i=1}^{n}\left(\int_{0}^{1}x^{i-1}dx\binom{n}{i}\right)=\int_{0}^{1}\sum_{i=1}^{n}\binom{n}{i}x^{i-1}dx$$
then I can't,Thank you
 A: I cannot find a closed form, but I can get a different sum which has an easier asymptotic expansion:
$$
\begin{align}
a_n
&=\sum_{i=1}^n\frac1i\binom{n}{i}\\
&=\sum_{i=1}^n\frac1i\left[\binom{n-1}{i}+\binom{n-1}{i-1}\right]\\
&=a_{n-1}+\sum_{i=1}^n\frac1n\binom{n}{i}\\
&=a_{n-1}+\frac{2^n-1}{n}
\end{align}
$$
Thus,
$$
\begin{align}
a_n
&=\sum_{i=1}^n\frac{2^i-1}{i}\\
&\sim\frac{2^{n+1}}{n}\left(1+\frac1n+\frac3{n^2}+\frac{13}{n^3}+\frac{75}{n^4}+\dots\right)
\end{align}
$$

Asymptotic Expansion
$$
\begin{align}
\frac{n}{2^{n+1}}\sum_{i=1}^n\frac{2^i-1}{i}
&=1+\frac{n}{2^{n+1}}\left[\sum_{i=1}^n\frac{2^i-1}{i}-\sum_{i=1}^n\frac{2^i}{n}-\frac2n\right]\\
&=1+\frac{n}{2^{n+1}}\left[\sum_{i=1}^n2^i\left(\frac1i-\frac1n\right)-\left(H_n+\frac2n\right)\right]\\
&=1+\sum_{i=1}^n2^{i-n}\frac{n-i}{2i}-\frac{n}{2^{n+1}}\left(H_n+\frac2n\right)\\
&=1+\sum_{i=0}^{n-1}2^{-i}\frac{i}{2(n-i)}\color{#C00000}{-\frac{n}{2^{n+1}}\left(H_n+\frac2n\right)}\\
&\sim1+\sum_{i=0}^\infty2^{-i}\frac12\left(\frac{i}{n}+\frac{i^2}{n^2}+\frac{i^3}{n^3}+\dots\right)\\
&\sim1+\frac1n+\frac3{n^2}+\frac{13}{n^3}+\frac{75}{n^4}+\dots
\end{align}
$$
The red term decays exponentially, so it can essentially be ignored.
A: Using your idea, sooner or later one reaches the formula
$$
f(n)=n\int_0^1\!\!\int_0^1(1+xy)^{n-1}\mathrm dx\mathrm dy.
$$
Whether this is a "closed form" or not is at best debatable...
A: There is a closed form which is probably not the most pleasant one since it involves an hypergeometric function 
$$f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}=n \text { } \, _3F_2(1,1,1-n;2,2;-1)$$ If $\log (f(n))$ is plotted as a function of $n$, it looks quite linear (may be this could give an idea to other participants).
A: $$\sum_{i=1}^n {n\choose i} x^{i-1} = \frac1x \left(\sum_{i=1}^n{n\choose i} 1^{n-i}x^{i}\right)=\frac1x\left(\sum_{i=0}^n{n\choose i} 1^{n-i}x^{i} -1\right)= \frac1x\left((1+x)^n - 1\right)$$
A: Just another way at ending at an integral form for the answer:
$$(1+x)^n = \sum_{i=0}^n \binom{n}{i}x^i \implies \sum_{i=1}^n \binom{n}{i}x^{i-1} = \frac{(1+x)^n-1}x$$
Now integrate both sides and set $x=1$. Mathematica gives a hypergeometric function as the answer, so I guess for now the best is:
$$ \left. \int_0^x \frac{(1+t)^n- 1}{t}dx \right |_{x=1}$$
