$\sum_{N=6}^{\infty}\frac{6}{N}\binom{N-1}{5}p^6(1-p)^{N-6}$ 
Find the sum: $\sum\limits_{N=6}^{\infty}\frac{6}{N}\binom{N-1}{5}p^6(1-p)^{N-6}$

I could not find a way to manipulate $\binom{N-1}{5}$ to get any suitable form here. Note that $\binom{N-1}{5}p^6(1-p)^{N-6}$ are probability masses of negative binomial distribution.

What we want to find:
$$\sum_{k=0}^\infty \frac{6}{k+6}\binom{k+5}{k}p^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$\sum_{k=0}^\infty \binom{k+5}{k}z^k =\frac{1}{(1-z)^6} $$
Some manipulation, and integration:
$$\int_0^{1-x}\sum_{k=0}^\infty \binom{k+5}{k}z^{k+5} dz =\sum_{k=0}^\infty \frac{1}{k+6}\binom{k+5}{k}(1-x)^{k+6} =  \int_0^{1-x} \frac{z^5}{(1-z)^6}dz = \frac{300x^4-300x^3+200x^2-75x+12}{60x^5}+\ln(x)-\frac{137}{60} $$
After substitution, and some multiplying by constants:
$$S(p) =\frac{300p^5-300p^4+200p^3-75p^2+12p}{10(1-p)^6}+\frac{6p^6\ln(p)}{(1-p)^6}-\frac{137p^6}{10(1-p)^6} $$
