Mean of a function of Binomial Distribution

Question

Let $X$ be a random variable following the Binomial Distribution with parameters $n$ and $p$. Show that $$ \mathbb{E}\left[\frac{1}{1+X}\right]=\frac{1-\left(1-p\right)^{n+1}}{p(n+1)}, $$ where $\mathbb{E}[\cdot]$ is the mean value funtion.

My textbook had this exercise and I found it very interesting, but i can't solve it because I suck at infinite sums with the binomial coefficient. Any suggestion is appreciated. Thank you very much in advance.

Answer 1

Note that $$ E\frac{1}{1+X}=\sum_{k=0}^n\frac{1}{1+k}\binom{n}{k}p^k(1-p)^{n-k}= \sum_{k=0}^n\left[\binom{n}{k}p^k(1-p)^{n-k}\int_{0}^1t^k\,dt\right] $$ i.e. $$ E\frac{1}{1+X}=\int_{0}^1\sum_{k=0}^n \binom{n}{k}(pt)^k(1-p)^{n-k}\, dt=\int_0^1(1-p+pt)^n\,d t $$ where we have used the linearlity of the integral and the binomial theorem. I leave it to you to compute the integral to yield the desired answer.