Trying to simplify $ \sum_{k=1}^n{n\choose k}\frac{1}{k^l}p^k(1-p)^{n-k}. $

Question

I am trying to simplify the following expression:

$$ \sum_{k=1}^n{n\choose k}\frac{1}{k^l}p^k(1-p)^{n-k}. $$

I am wondering if I can simplify the sum away where $l\in\{1,2\}$. Thank you!

Answer 1

For $l=1$: take the lower bound $\frac{1}{k} > \frac{1}{k+1}$, multiply by $\frac{p(n+1)}{p(n+1)}$ $$ S_{n+1} \geq \frac{1}{p(n+1)}\sum_{k=0}^{n}\binom{n+1}{k+1} p^{k+1}(1-p)^{n-k} = \frac{1}{p(n+1)}\bigg(1-(1-p)^{n+1}\bigg) $$

Answer 2

Both Maple and Wolfram Alpha express the sums using generalized hypergeometric functions: for $l=1$, $$ np \left( 1-p \right) ^{-1+n}\; {\mbox{$_3$F$_2$}(1,1,1-n;\,2,2;\,{\frac {p}{-1+p}})} $$ and for $l=2$, $$np \left( 1-p \right) ^{-1+n}\; {\mbox{$_4$F$_3$}(1,1,1,1-n;\,2,2,2;\,{\frac {p}{-1+p}})} $$ It's possible there might be a simpler form, but I wouldn't hold my breath.

EDIT: If your expression is $S_l(n)$, the generating function of $S_1$ seems to be

$$ g_1(x) = \sum_{n=0}^\infty S_1(n) x^n = \frac{\ln(1-(1-p)x) - \ln(1-x)}{1-(1-p)x} $$