# Sum involving binomial coefficients

I have the following sum $\sum_{j=1}^K {K \choose j} (-1)^{j+1}/j$. Now I can write this as the integral $\int_{-1}^0 \frac{(1+x)^K - 1}{x} dx$. However, I wonder whether there is a closed form expression for that integral? Thanks.

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Let $u=x+1$. $\int_{-1}^0 \frac{(1+x)^K-1}{x}dx = \int_0^1 \frac{u^K-1}{u-1}du = \int_0^1 (u^{K-1} + u^{K-2} + ... + u + 1) du$. The result is the $K$th harmonic number $H_K = 1/K + 1/(K-1) + ... + 1/2 + 1$. –  Douglas Zare Jan 19 '14 at 6:36

## 2 Answers

The closed form of the integral is

$$\int_{-1}^{0}\frac{(1+x)^k - 1}{x} = \frac{s(k+1,2)}{k!}$$

where $s(k+1,2)$ denote the Stirling number of the first kind.

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Is there a similar integral representation for $s(n,r)$ ? –  Pietro Majer Feb 4 '13 at 12:32
Are you sure about this? Then this would just reduce to $\sum_{j=1}^k 1/k$, right? –  Danne Feb 4 '13 at 14:20
I think one has to involve Stirling functions of the second kind, not Stirling numbers of the first kind. –  Danne Feb 4 '13 at 14:36
@Pietro Yes there is such an integral which is in fact a contour integral. $$s(n,r) = \frac{n!}{2\pi r!}\int_{|z|=1} z^{-n-1} \log^r (z+1)dz$$ @Danne, Yes I am sure. –  Nilotpal Sinha Feb 4 '13 at 16:10
But when I evalute that integral and the Stirling formula you gave, I don't get the same answer. How do you find that formula? –  Danne Feb 4 '13 at 16:15


$$\begin{array}{rcl} {\rm B}\pars{\alpha,\beta}=\int_{0}^{1}t^{\alpha - 1}\pars{1 - t}^{\beta - 1}\,\dd t: && Beta\ \mbox{function} \\[2mm] \Gamma\pars{z}:&& Gamma\ \mbox{function} \\[2mm] \Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}:&& Digamma\ \mbox{function} \\[2mm]\gamma \approx 0.577216:&& Euler-Mascheroni\ constant \end{array}$$
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