Question : Let $X$ be a binomial random variable with parameters $n$ and $0<p<1$.
- Find $\mathbb{E}\Big(\frac{1}{1+X}\Big)$
- Show that $\lim\limits_{n\to\infty} \mathbb{E}(1+X)\cdot \mathbb{E}\Big(\frac{1}{1+X}\Big)=1$
My Attempt : I have already computed $\mathbb{E}\Big(\frac{1}{1+X}\Big)$. But while solving another part, I reached here: $$\lim\limits_{n\to\infty}\bigg((np+1)\cdot\frac{(1-(1-p)^n+1)}{(n+1)p}\bigg)$$
I got confused. What steps should I take to compute the answer?
You should have something that looks like this: $$ \lim_{n \to\infty} (1 + np)\cdot\left(1 - (1-p)^{n+1} \over (n+1)p \right). $$
When we distribute all of this through we will end up with:
$$ \lim_{n \to\infty} \left(1 - (1-p)^{n+1} \over (n+1)p \right) + \lim_{n \to\infty} \left(np - np(1-p)^{n+1} \over (n+1)p \right). $$
In the first equation, the numerator will never exceed $2$ because $(1-p)^{n+1}$ is going to be less than $1$ because we know that $0 \lt p \lt 1.$ but the denominator is going to blow up to $\infty$, making the first term completely $0$.
On the term on the right, I find it easiest to artfully multiply by 1, by dividing both the numerator and denominator by $\frac{1}{n}$, but taking L'Hopital's would also work. Observe:
$$ \left(\frac{1}{n}[np - np(1-p)^{n+1}] \over \frac{1}{n}[(n+1)p] \right) = \left( p - p(1-p)^{n+1} \over p + \frac{p}{n} \right). $$
Under the limit the $p(1-p)^{n+1}$ is going to go to zero, as well as the $\frac{p}{n}$, and you will be left with $\frac{p}{p}$ which equals $1$ as desired.