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Is there a closed form of the series:

$$\sum_{k=0}^{n}\frac{\binom{n}{k}}{nk+1}?$$

Thank you in advance for any help.

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    where did this come from? and why do you think there is a closed form? – 2017-02-27
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    The series appears, when I try to solve the integral int (1+x^n)^n with limits 0 and 1 – 2017-02-27
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    Give the integral to ensure we are solving the correct thing (and gives the question more context :) ) – 2017-02-27
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    The look of this series made me want to transform it into that very integral. I mean, it is no easier to handle this way than that. – 2017-02-27
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    [Wolfram alpha gives](http://www.wolframalpha.com/input/?i=Sum%5BBinomial%5Bn,k%5D*(n*k%2B1)%5E(-1),%7Bk%3D0,n%7D%5D) an hypergeometric function as an answer. What I dont know is if this hypergeometric have a closed/simpler form. – 2017-02-27

2 Answers 2

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That is the integral over $(0,1)$ of $$ \sum_{k=0}^{n}\binom{n}{k}(x^n)^k = (1+x^n)^n \tag{1}$$ hence $$ \sum_{k=0}^{n}\binom{n}{k}\frac{1}{kn+1} = \int_{0}^{1}(1+x^n)^n\,dx=\frac{1}{n}\int_{0}^{1}z^{1/n-1}(1+z)^n\,dz \tag{2}$$ giving the hypergeometric function shown in the other answer.
By Laplace's method, the LHS of $(2)$ is expected to behave like $\frac{2^{n+1}}{n^2}$.

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$$\sum_{k=0}^{n}\frac{\binom{n}{k}}{nk+1}=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{nk+1}=_2F_1(\frac{1}{n}, -n;\frac{1}{n}+1;-1)$$

It is the Hypergeometric function