I need to bring the following formula into a form that is more easy to handle:
$$\sum_{l=0}^k\frac{1}{l+1}B_{k,p}(l)=\sum_{l=0}^k\frac{1}{l+1}{k \choose l}p^l(1-p)^{k-l}$$
Luckily, Wolfram Alpha tells me that it holds that:
$$\sum_{l=0}^k\frac{1}{l+1}{k \choose l}p^l(1-p)^{k-l}=\frac{p(1-p)^{k}-(1-p)^k + 1}{kp+p}$$
I'm trying to understand how Wolfram Alpha comes to that conclusion, but I'm unable to transform the left-hand side into the right-hand side. What are the steps to transform the original term into the more succinct form?
Note that $\frac{1}{l+1}\binom{k}{l} = \frac{1}{k+1}\binom{k+1}{l+1}$. Thus \begin{align} \sum_{l = 0}^k \frac{1}{l+1}\binom{k}{l}p^l(1-p)^{k-l} &= \frac{1}{k+1}\sum_{l = 0}^k \binom{k+1}{l+1}p^{l+1-1}(1-p)^{(k+1) - (l+1)} \\ &= \frac{1}{(k+1)p} \sum_{r = 1}^{k+1} \binom{k+1}{r} p^r (1-p)^{k+1-r} \\ &= \frac{1}{(k+1)p} \bigl((p + 1-p)^{k+1} - (1-p)^{k+1}\bigr) \\ &= \frac{1 - (1-p)^{k+1}}{(k+1)p}\,. \end{align} You can manipulate the numerator to obtain the form Wolfram Alpha put out by distributing $$(1-p)^{k+1} = (1-p)(1-p)^k = (1-p)^k - p(1-p)^k$$ if you wish.