Closed form of $\sum_{k=1}^{n} \frac{{n}\choose{k}}{k}$.

Question

I would like to ask if it is possible to find a closed form of the sum $\sum_{k=1}^{n} \frac{{n}\choose{k}}{k}$ (1).
I managed to show that it is enough to find a closed form of $\sum_{k=1}^{n} \frac{2^k}{k}$ which is equivalent to computing the integral $\int_{1}^{2}\frac{x^n-1}{x-1}dx$.

Question: can we find a closed form of (1) in terms of elementary (Polynomial, exponential) functions?

Answer 1

$\displaystyle (1+x)^n = \sum_{k = 0}^{n} {n \choose k} x^k$

$\displaystyle \frac{(1+x)^n}{x} = \sum_{k = 0}^{n} {n \choose k} x^{k-1}$

$\displaystyle \int \frac{(1+x)^n}{x} dx = log(x) + \sum_{k = 1}^{n} {n \choose k} \frac{x^{k}}{k}$

$\displaystyle \sum_{k = 1}^{n} {n \choose k} \frac{x^{k}}{k} = -log(x) + \int \frac{(1+x)^n}{x} dx$

When $x = 1$ , $log(1) = 0$

From wolfram

$\displaystyle \int \frac{(1 + x)^n}{x} dx = \frac{(\frac{1}{x} + 1)^{-n} (x + 1)^n \: _2F_1(-n, -n; 1 - n; -\frac{1}{x})}{n} + constant $

$_2F_1$ : Hypergeometric Function

$\displaystyle \sum_{k = 1}^{n} \frac{1}{k} {n \choose k} = \frac{_2F_1(-n, -n; 1 - n; -1)}{n} + constant $

No its not an elementary function which implies it cannot be (sometimes simplifications exist).

Wolfram even simpler

sum nchoosek(n,k)/k , k = 1 to n

$\displaystyle \sum_{k=1}^n \frac{1}{k}{n \choose k} = n \: _3 F_2(1, 1, 1 - n;2, 2;-1)$

Generalized Hypergeometric Function