Evaluate $$\lim_{n\rightarrow\infty}\sum_{r=0}^{n}\dfrac{{n\choose r}}{n^r\cdot(r+3)}$$
This form forcing me to use integrals, I tried expanding $${n\choose r}=\dfrac{n(n-1)\cdots (n-r+1)}{r!}$$ then dirtributing that $n^r$ to each braket to generate $$\dfrac{1(1-\frac{1}{n})\cdots (1-\frac{r-1}{n})}{r!}$$ then I don't know what to do with $r!$ and $r+3$.
How to proceed further?
$$\lim_{n\rightarrow\infty}\sum_{r=0}^{n}\dfrac{{n\choose r}}{n^r\cdot(r+3)}=\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{\binom{n}{r}}{n^r}\int^{1}_{0}x^{r+2}dx$$
$$=\int^{1}_{0}x^2\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\binom{n}{r}\bigg(\frac{x}{n}\bigg)^rdx$$
$$=\int^{1}_{0}x^2\lim_{n\rightarrow \infty}\bigg(1+\frac{x}{n}\bigg)^{n}dx=\int^{1}_{0}x^2\cdot e^{x}dx$$
More generally, for any positive integer $k$, $$\sum_{r=0}^{n}\dfrac{{n\choose r}}{n^r\cdot(r+k)}= n^k\int_0^{1/n}x^{k-1}(1+x)^n \, dx =\int_0^{1}t^{k-1}\left(1+\frac{t}{n}\right)^n dt\to \int_0^{1}t^{k-1}e^t \, dt$$ where $t=nx$ and at the last step we used Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
