WolframAlpha can't calculate this limit.

Truth be told, it took me twice attempts to answer this problem and I really wonder how members on Brilliant answered this problem.

This is my approach:

When I saw this problem, it reminded me with the probability mass function (PMF) of Poisson distribution with parameter $n$ and in statistics theory, sum of all supports of the PMF is equal to $1$ . So my first answer was $2$ by considering $a=1$ and $b=1$ . But as you can see, my answer was incorrect. I can still use more rigor statistics approach by using Central Limit Theorem (CLT) as $n\to\infty$ , the limiting distribution of Poisson distribution will be normal distribution with mean $n$ and variance $n$ , but I prefer using another approach and I think it is simpler.

If we expand the series and take the limit of each term, it is clearly all terms but the first term are equal to $0$ .

Proof: $$ $\begin{aligned} \lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!} &=\lim_{n\to\infty} e^{-n}\left(\lim_{k\to0}\frac{n^k}{k!}+\frac{n^1}{1!}+\frac{n^2}{2!}+\cdots\right)\\ &=\lim_{n\to\infty}\lim_{k\to0}\frac{n^k}{k!\,e^{n}}+\lim_{n\to\infty} \frac{n}{e^{n}}+\lim_{n\to\infty} \frac{n^2}{2\,e^{n}}+\cdots \end{aligned}$ $$ The first term is indeterminate form of $\infty^0$ in the numerator part, therefore it needs special treatment.

The $k$ -term for $k\ge2$ where $k$ is finite, it is clearly the limit is $0$ since the exponential function (the denominator part) goes to infinity faster than the polynomial function (the numerator part).

The $n$ -term where $n\to\infty$ , the limit is $$ $\lim_{n\to\infty} \frac{n^n}{n!\,e^{n}}.$ $$ As $n\to\infty$ , we can use Stirling's formula $$ $n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$ $$ Hence, $$ $\begin{aligned} \lim_{n\to\infty}\frac{n^n}{n!\,e^{n}} &=\lim_{n\to\infty}\frac{n^n}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\cdot e^{n}}\\ &=\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\\ &=0.\quad\quad\quad\quad\quad\blacksquare \end{aligned}$ $$ Since the first term is indeterminate form, we will use another simple statistics approach. Let $X$ be the value of $\lim\limits_{n\to\infty}\lim\limits_{k\to0}\;n^k$ . It is obvious that $X$ is a random variable. I claim that the value of $X$ lies within the interval $k<X<e^n$ , where $k\to0$ and $n\to\infty$ . It is reasonable if we assume that the distribution of $X$ is continuous uniform distribution since each value of $X$ in $(k,e^n)$ is equally probable. In statistical term, $X\sim\mathcal{U}(k,e^n)$ . The expectation value or mean of $X$ is easy to be obtained when we deal with uniform distribution. $$ $\text{E}[X]=\frac{1}{2}(k+e^n).$ $$ Therefore $$ $\begin{aligned} \lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!} &=\lim_{n\to\infty}\lim_{k\to0}\frac{n^k}{k!\,e^{n}}\\ &=\frac{\lim\limits_{n\to\infty}\lim\limits_{k\to0}\;n^k}{\lim\limits_{n\to\infty}\lim\limits_{k\to0}\;k!\,e^{n}}\\ &=\frac{\lim\limits_{n\to\infty}\lim\limits_{k\to0}\;\frac{1}{2}(k+e^n)}{\lim\limits_{n\to\infty}\lim\limits_{k\to0}\;k!\,e^n}\\ &=\frac{1}{2}. \end{aligned}$ $$ Thus, $a+b=1+2=\boxed{\color{#3D99F6}{3}}$ . $$

P.S. : For other viewpoint to approach this problem, you may want to download and read this journal: On The Limit of A Sequence . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$