Suppose that a random variable $X$ can take $5$ values $\{1,2,3,4,5\}$ with probabilities that depend upon $n \geq 0$ and are given by
\[P(X=k)=\frac{e^{k n}}{e^{n}+e^{2 n}+e^{3 n}+e^{4 n}+e^{5 n}}\]
for $k=1,2,3,4,5$. What can one say about the expectation $E[X]$ as $n \rightarrow \infty$?
- It increases to infinity as $n \rightarrow \infty$
- It equals $3$ for all values of $n \geq 0$
- It converges to $1$ as $n \rightarrow \infty$
- It converges to $5$ as $n \rightarrow \infty$
- It converges to $0$ as $n \rightarrow \infty$