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I wanted to compute the sum $$\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}.$$ And I thought it would be easiest to do this by making it a function, differentiating it and integrating it then.

So I did: $$f_n(a)=\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}a^k$$ $$f_n'(a)=\sum_{k=1}^{n}\binom{n}{k}a^{k-1}=\frac{(a+1)^n-1}{a}$$ And finally $$f_n(a)=\int\frac{(a+1)^n-1}{a}\mathrm{d}a+C(n)$$ where $C(n)$ is an unknown constant depending on $n$.

The first problem is the integral of course, so here it stops.

Can anyone help me calculating this integral or is there another way to calculate the sum?

EDIT: another approach:

I get, using Pascal's identity, $$S(n)=\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}=\frac{1}{n}+\sum_{k=1}^{n-1}\left(\frac{1}{k}\binom{n-1}{k-1}+\frac{1}{k}\binom{n-1}{k}\right)=\sum_{k=1}^{n}\frac{1}{n}\binom{n}{k}+\sum_{k=1}^{n-1}\frac{1}{k}\binom{n-1}{k}$$

So I have the recursion $S(n)=\frac{2^n-1}{n}+S(n-1)$, with $S(1)=\frac{2^1-1}{1}$.

Thus $$\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}=\sum_{k=1}^{n}\frac{2^k-1}{k}$$ if that might help.

Also, $f_n(a)=\displaystyle\sum_{k=1}^{n}\frac{(a+1)^k-1}{k}$ for the same reason.

Bart Michels
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1 Answers1

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Here is a formula for the sum in terms of the hypergeometric function

$$\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}= n\, _3F_2(1,1,1-n;\,2,2;\,-1).$$

$$\,_3F_2(a_1,a_2,a_3;b_1,b_2;z) = \sum_{n=0}^\infty {(a_1)_n (a_2)_n (a_3)_n\over (b_1)_n (b_2)_n } \, {z^n \over n!} $$

Mhenni Benghorbal
  • 47,431