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Context:

From [1,2,3], one can obtain identities for the following

$ { \sum\limits_{k=1}^N{ \binom{N}{k}\, k^r\, } } $, where $r$ is a positive integer and $\binom{N}{k}$ are binomial coefficients.

Question 1.

Is there an identity for

$ { \sum\limits_{k=1}^N{ \binom{N}{k}\,\dfrac{p^k \, \left( - 1 \right)^{k} }{k} } } $, where $0< p < 1$?

Question 2.

If no such identity is known, then the next best thing would be to have an upper and a lower bound. For example, its clear that

$ { \sum\limits_{k=1}^N{ \binom{N}{k}\,\dfrac{p^k \, \left( - 1 \right)^{k} }{k} } } < { \sum\limits_{k=1}^N{ \binom{N}{k}\, p^k \, \left( - 1 \right)^{k} } } $ for $0< p < 1$.

Are there suggestions for expressions that together tightly bind the fintite series from above and below?

References:

[1] Boros and Moll 2004, pp. 14-15

[2] http://mathworld.wolfram.com/BinomialSums.html

[3] https://en.wikipedia.org/wiki/Binomial_coefficient#Series_involving_binomial_coefficients

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In case this helps:

$$\frac{a^k}{k}= \int_0^a x^{k-1} dx \hskip{1cm} k>0$$

Hence

$$ \sum_{k=1}^n {n \choose k} \frac{a^k}{k} = \int_0^a \frac{(1+x)^n-1}{x} dx$$

Or

$$ \sum_{k=1}^n {n \choose k} (-1)^k\frac{p^k}{k} = \int_0^p \frac{(1-u)^n-1}{u} du$$

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  • $\begingroup$ While I appreciate the comment of leonbloy, my problem actually began with the integral that leonbloy ends up with. So, I suppose that a bounding (see Question 2 above) is in order. $\endgroup$ – Michael Levy Dec 2 '16 at 14:58

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