Is the summation $\sum_{i=1}^{n}\frac1{i} \binom{n}{i}$ possible? I want to compute the following sum:
$$
\sum\limits_{i=1}^{n} \frac{{n\choose{i}}}{i}
$$

What I have done so far:
We know that $$(1+x)^n=\sum\limits_{r=0}^{n} {n\choose{r}}x^r$$
so, $$\frac{(1+x)^n-1}{x}=\sum\limits_{i=1}^{n} {{n\choose{i}}}x^{i-1}$$
therefore, upon integration we get,
$$\int\limits_{0}^{1}\frac{(1+x)^n-1}{x}dx=\sum\limits_{i=1}^{n} \frac{{{n\choose{i}}}}{i}$$
I cannot get any further with the LHS of the above equation.

Primary questions to be addressed:

*

*Is such an integration possible (why so)?

*Are there any other approximations for the sum?

 A: To expand a little on @Sangchul Lee's comment - after the substitution of ${y=1+x}$ the integral becomes
$${\Rightarrow \int_{1}^{2}\frac{y^n-1}{y-1}dy}$$
(This substitution isn't "necessary" for the next part, however it makes it a bit clearer). Now we can use a special factoring formula:
$${a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})}$$
To get
$${\int_{1}^{2}\frac{(y-1)(y^{n-1} + y^{n-2} + ... + y + 1)}{(y-1)}dy=\int_{1}^{2}y^{n-1} + y^{n-2} + ... + y + 1dy}$$
Evaluating that last integral gives us the sum
$${\Rightarrow \sum_{k=1}^{n}\frac{2^{k}-1}{k}}$$
So overall you have that
$${\sum_{k=1}^{n}\frac{{n\choose k}}{k}=\sum_{k=1}^{n}\frac{2^k-1}{k}}$$
Other than that though, I'm not sure if there's any more useful form. WolframAlpha gives some very nasty looking closed forms involving special functions.

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{i = 1}^{n}{{n \choose i} \over i} & =
\sum_{i = 1}^{n}{n \choose i}\int_{0}^{1}t^{i - 1}\,\dd t =
\int_{0}^{1}\sum_{i = 1}^{n}{n \choose i}t^{i}\,{\dd t \over t} =
\int_{0}^{1}\bracks{\pars{1 + t}^{n} - 1}\,{\dd t \over t}
\\[5mm] & =
\int_{1}^{2}{1 - t^{n} \over 1 - t}\,\dd t =
\int_{0}^{2}{1 - t^{n} \over 1 - t}\,\dd t\ -\
\underbrace{\int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t}_{\ds{H_{n}}}
\\[5mm] & =
\int_{0}^{2}{1 - t^{n} \over 1 - t}\,\dd t
- H_{n}
\end{align}
