How can I possibly calculate
$$\sum_{k=1}^{\infty} k {n+k \choose n} p^k$$ ???$$
I tried naming that as $U_n$ and used the fact that the sum is infinite and that
$$^{n}C_p = ^{n-1}C_p + ^{n-1}C_{p-1}$$
But I arrive at $$U_n = pU_n + U_{n-1} + T_n$$
Where $$T_n = \sum_{k=1}^{\infty} {{n+k-1} \choose n} p^k$$ another sum I can't calculate
So I think my approach is wrong Any help would be appreciated Thanks in advance!!
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$\sum\limits_{k=0}^{\infty} k {n+k \choose k} p^k=p \frac{d}{dp}\sum\limits_{k=0}^n p^k {n+k \choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$\sum\limits_{k=0}^n p^k {n+k \choose k}=\frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$\sum\limits_{k=1}^{\infty} k {n+k \choose n} p^k=\frac {p(n+1)}{(1-p)^{(n+2)}}$