Usually I would write the given sum in the form $$lim_{n \to \infty}\frac{1}{n}\sum_{r=o}^{n}{f(\frac{r}{n})}$$ and then approximate it with the integral
$$\int_{0}^{1}f(x)dx$$ but it doesn't seem so easy to do with this question. The solution says that this sum is equal to the integral: $$\int_{0}^{1}x^2e^xdx$$ without any further explanation. I can't see how they are equal.
Any help is appreciated.
$$\lim_{n\to \infty} \sum_{k=0}^n {n \choose k} \frac{1}{(k+3)n^k} = \lim_{n\to \infty} \sum_{k=0}^n {n \choose k} \frac{1}{n^k} \int_0^1 x^{k+2}\:dx $$
$$\int_0^1 x^2 \lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n\:dx = \int_0^1 x^2 e^x \:dx $$
with appropriate assumptions on uniform convergence.
Just for the fun of it !
Since you already received @Ninad Munshi's good answer, let us try to look at the partial sums $$S_n=\sum_{k=0}^{n} \frac{{n\choose k}}{n^k(k+3)}$$ What it could "easily" be shown is that $$S_n=\frac{\left(1+\frac{1}{n}\right)^n (n+1) \left(n^2+n+2\right)-2 n^3}{(n+1) (n+2) (n+3)}$$ Using series for large values of $n$ $$S_n=(e-2)+\frac{24-9 e}{2n}+\frac{443 e-1200}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
Trying for $n=10$, the exact value is $$S_{10}=\frac{249060564759}{357500000000}\approx 0.696673$$ while the above truncated expansion gives $$\frac{1763 e}{2400}-\frac{13}{10}\approx 0.696805$$
