How to find this limit using properties of Poisson variables? Let $X$ and $Y$ be two independent Poisson random variables, with means $\lambda_1$ and $\lambda_2.$ Then $X + Y$ is a Poisson random variable with mean $\lambda_1+ \lambda_2.$ Arguing in a similar way, a Poisson random variable $X$ with parameter $t,$ where $t$ is a positive integer, can be thought of as sum of $t$ independent Poisson random variables $X_1, X_2,...,X_t$, each of which has mean $1.$
Using the information above, and an appropriate limit theorem, evaluate the following limit:
$$\lim _{n \rightarrow \infty} \sum_{k>n+\sqrt n}^\infty \frac{e^{-n}n^k}{k!}$$
I tried to evaluate the limit, but it doesn't seem to work. I have also tried to do maximum likelihood, but it doesn't work either.
 A: Let $X_i,$ $i=1,2,3,\ldots,$ be a sequence of independent Poisson$(1)$ variables.  Recall that when a random variable $X$ has a Poisson$(\lambda)$ distribution and $k\in\{0,1,2,\ldots\}$ (the natural numbers),
$$\Pr(X = k) = e^{-\lambda}\, \frac{\lambda^k}{k!}.$$
In the sequel, we will use the facts that both the expectation and variance of $X$ equal $\lambda.$
Here are the two observations made in the question:


*

*$Y_n = X_1 + X_2 + \cdots + X_n$ has a Poisson$(n)$ distribution.  Thus, for every natural number $k,$ $$\Pr(Y_n = k) = e^{-n}\, \frac{n^k}{k!}.$$

*The Central Limit Theorem implies the distribution of $Z_n = (Y_n - E[Y_n]) / \sqrt{\operatorname{Var}(Y_n)}$ converges to the standard Normal distribution (whose CDF I will call $\Phi$).
Writing $F_n$ for the CDF of $Z_n,$ (2) asserts that for every number $x$ where $\Phi$ is continuous (which is all real numbers),
$$\lim_{n\to\infty} F_n(x) = \Phi(x).$$
Equivalently,
$$\lim_{n\to\infty} 1 - F_n(x) = 1 - \Phi(x) = \Phi(-x).$$
Notice two things: first, $$Y_n = Z_n \sqrt{\operatorname{Var}{Z_n}} + E[Z_n] = Z_n \sqrt{n} + n$$ and second, in light of that result the event $Z_n \gt x$ can equivalently be described as $$\{Z_n \gt x \} = \{Z_n \sqrt{n} + n \gt n + x\sqrt{n}\}= \{Y_n \gt n + x\sqrt{n}\}.$$
Consider $x=1$ and use the foregoing facts to compute
$$\eqalign{
1 - F_n(1) &= 1 - \Pr(Z_n \le 1) = \Pr(Z_n \gt 1)\\
&= \Pr(Y_n \gt n + \sqrt{n}) \\
&= \sum_{k \gt n + \sqrt{n}} \Pr(Y_n=k) \\
&= \sum_{k \gt n + \sqrt{n}} e^{-n}\,\frac{n^k}{k!}
.}$$
Therefore

$$\lim_{n\to\infty}\, \sum_{k \gt n + \sqrt{n}} e^{-n}\,\frac{n^k}{k!} = \lim_{n\to\infty}(1 - F_n(1)) = \Phi(-1) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-1} e^{-x^2/2}\,\mathrm{d}t.$$

