Limit of $\sum\limits_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}$ when $n\to\infty$ What is $$ \lim_{n \to  \infty} \sum_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}\ ?$$ I know the way by integration and that the answer is $e-2$ but I am more interested in use of sandwich theorem which provides a maxima or a closed form to it. Expansions may also be useful.

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Besides the
  'sandwich' answer of $\texttt{@RobertZ}$,
  the following answer provides an alternative approach:

\begin{align}
\lim_{n \to  \infty}\sum_{k = 0}^{n}{{n \choose k} \over n^{k}\pars{k + 3}} & =
\lim_{n \to  \infty}\sum_{k = 0}^{n}{n \choose k}
\pars{1 \over n}^{k}\int_{0}^{1}t^{k + 2}\,\dd t =
\lim_{n \to  \infty}\int_{0}^{1}t^{2}\sum_{k = 0}^{n}{n \choose k}
\pars{t \over n}^{k}\,\dd t
\\[5mm] & =
\lim_{n \to  \infty}\int_{0}^{1}t^{2}\pars{1 + {t \over n}}^{n}\,\dd t =
\lim_{n \to  \infty}\bracks{n^{3}\int_{0}^{1/n}t^{2}\pars{1 + t}^{n}\,\dd t}
\end{align}

The last integral can be evaluated by succesive integration by parts which decreases the $\ds{t^{2}\mbox{-exponent}}$ to $\ds{0}$.

Namely,
\begin{align}
&\lim_{n \to  \infty}\sum_{k = 0}^{n}{{n \choose k} \over n^{k}\pars{k + 3}}
\\[5mm] = &\
\lim_{n \to \infty}\braces{n^{3}\bracks{-2n^{3} + \pars{1 + 1/n}^{n}\pars{1 + n}\pars{2 + n + n^{2}} \over n^{3}\pars{1 + n}\pars{2 + n}\pars{3 + n}}} =
\bbx{\expo{} - 2}
\end{align}
