Is the limit $ \lim_{n\to \infty}\left(\sum^{n}_{r=0} \binom{n}{r}\big/{n^{r}(r+3)}\right)$ rational or irrational? How can I prove that the result of the following limit is rational/irrational?$$ \lim_{n\to \infty}\left(\sum^{n}_{r=0} \frac{\binom{n}{r}}{n^{r}(r+3)}\right)$$
Would solving this limit satisfy? How would I solve this? So, if the result came out to be say $\pi$, can we conclude that it is irrational? Meaning, is it a complete proof?
 A: Let $f_n(x)=\sum_{r=0}^n\,\binom{n}{r}\,\frac{x^r}{r+3}$ for $x\in\mathbb{R}$ and $n\in\mathbb{N}$.  Then, 
$$\frac{\text{d}}{\text{d}x}\,\left(x^3\,f_n(x)\right)=\sum_{r=0}^n\,\binom{n}{r}\,x^{r+2}=x^2(1+x)^n\,.$$
Because $x^3f_n(x)$ is $0$ when $x=0$, we conclude that
$$f_n(x)=x^{-3}\,\int_0^x\,t^2(1+t)^n\,\text{d}t\,.$$
Write $a_n:=f_n\left(\frac{1}{n}\right)$.  We are looking for $\displaystyle\lim_{n\to\infty}\,a_n$.  
Note that
$$a_n=n^3\,\int_0^{1/n}\,t^2(1+t)^n\,\text{d}t=\int_0^1\,s^2\left(1+\frac{s}{n}\right)^n\,\text{d}s\,,$$
where $s:=nt$.  Let $g_n(s):=s^2\left(1+\frac{s}{n}\right)^n$ for $n\in\mathbb{N}$ and $s\in[0,1]$.  Observe that $g_n$ converges uniformly (or increasingly) to $g$ as $n\to\infty$, where $g(s):=s^2\exp(s)$ for all $s\in[0,1]$.  Hence, we can switch the limit and the integral, and obtain
$$\lim_{n\to\infty}\,a_n=\int_0^1\,s^2\exp(s)\,\text{d}s=\text{e}-2\,,$$
which is irrational.
In general, if $F_n(a,x):=\sum_{r=0}^n\,\binom{n}{r}\,\frac{x^r}{n^r(r+a)}$ for $n\in\mathbb{N}$, $a\in\mathbb{C}$ with $\text{Re}(a)>0$, and $x\in\mathbb{C}$, then $$F_n(a,x)=\int_0^x\,s^{a-1}\,\left(1+\frac{s}{n}\right)^n\,\text{d}s\,.$$
Hence,
$$\lim_{n\to\infty}\,F_n(a,x)=\int_0^x\,s^{a-1}\,\exp(s)\,\text{d}s=(-1)^{-a}\,\gamma(a,-x)\,,$$
where $\gamma$ is the lower incomplete gamma function.
A: $$\begin{equation} \begin{split} 
 \lim_{n \to \infty} { \sum_{r=0}^n {\frac{\left( _r^n \right)}{n^r(r+3)}}}  
&
 = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{n!}{r!(n-r)!n^r(r+3)}}} \\ 
&
 = \lim_{n \to \infty} { \sum_{r=0}^n {\left[  {\lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right)} \left( \frac{1}{r!(r+3)} \right)\right]}} \\ 
&
 = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{1}{r!(r+3)}}} \quad \quad \color{green} {\text{Note: } \lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right) = 1} \\ 
&
 = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{{x^{r+3}}}{r!(r+3)}}} \quad \quad \color{green}{\text{Note: For } x = 1} \\ 
&
 = \lim_{n \to \infty} { \sum_{r=0}^n { \int_0^1 {\frac{x^{r+2}}{r!}dx}}} \\ 
&
 = \int_0^1 x^2 \lim_{n \to \infty} \sum_{r=0}^n {\frac{x^r}{r!}} dx \\ 
&
 = \int_0^1 {x^2e^x \space dx} \\ 
&
 = \left[x^2e^x\right]_0^1 - 2 \int_0^1 {xe^x \space dx} \\ 
&
 = e - 2 [e-(e-1)] \\ 
&
 = e - 2 \end{split} \end{equation} $$
$\therefore \text{We can successfully conclude that}  \displaystyle\lim_{n \to \infty} { \sum_{r=0}^n {\frac{\left( _r^n \right)}{n^r(r+3)}}} \text{ is irrational.} 
$
