Evalute $ \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} $ 
Evaluate $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)}$.

$\bf{My\; Try::}$ Although we can solve it by converting into definite Integration.
But I want to solve it without Using Integration.
So $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{(n-1)(n-2).......(n-k+1)}{k!\cdot n^{k}\cdot (k+3)}$
Now How can i solve after that, Help required, Thanks
 A: One can do this without appealing to the DCT, essentially rendering the problem a lot more elementary, if complicated. All the steps below are justified because they're performed over a finite sum.
Start by rewriting the sum as follows
$$f(n):= \sum_{k=0}^n \binom nk \frac{n^{-k}}{k+3} = n^3 \sum_{k=0}^n \binom nk \frac{n^{-(k+3)}}{k+3} =: n^3 g(n) \tag{1}$$
Where
$$ g(x) := \sum_{k=0}^n \binom nk \frac{x^{-(k+3)}}{k+3} = \int_{x}^\infty \sum_{k=0}^n \binom nk y^{-(k+4)} \mathrm{d}y = \int_x^\infty \left(1 + y^{-1}\right)^n y^{-4}\mathrm{d} y$$
The substitution $y^{-1} = t$ and a bit of integration by parts yields
$$g(x) = \frac{x^{-2}\left(1+x^{-1}\right)^{n+1}}{n+1} -\frac{2x^{-1}\left(1+x^{-1}\right)^{n+2}}{(n+1)(n+2)} + \frac{2\left(1+x^{-1}\right)^{n+3}}{(n+1)(n+2)(n+3)} - \frac{2}{(n+1)(n+2)(n+3)} \tag{2}$$
Feeding $(2)$ into $(1)$ and taking the limit gives us
$$\lim_{n \to \infty} f(n) = e-2$$
(Which agrees with the answer by Kelenner, along with giving you an expression for $f(n)$ for arbitrary $n$)
A: This is not an answer but it is too long for a comment.
What I found interesting is that closed form expressions can be obtained for the partial sums since
$$S^{(j)}_n=\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+j)}=\frac{\, _2F_1\left(j,-n;j+1;-\frac{1}{n}\right)}{j}$$ and from there, the corresponding limits and asymptotics.
For the case where $j=3$ as in the post $$S^{(3)}_n=\frac{\left(1+\frac{1}{n}\right)^n (n+1) \left(n^2+n+2\right)-2 n^3}{(n+1) (n+2)
   (n+3)}=(e-2)+\frac{12-\frac{9}{2}e}{n}+O\left(\frac{1}{n^2}\right)$$ Similarly $$S^{(2)}_n=\frac{n^2+(n+1) \left(1+\frac{1}{n}\right)^n}{(n+1) (n+2)}=1+\frac{e-3}{n}+O\left(\frac{1}{n^2}\right)$$  $$S^{(4)}_n=\frac{6 n^4+\left(n \left(n \left(-2 n^2+n+8\right)+11\right)+6\right)
   \left(1+\frac{1}{n}\right)^n}{(n+1) (n+2) (n+3) (n+4)}=(6-2 e)+\frac{22 e-60}{n}+O\left(\frac{1}{n^2}\right)$$ where appear interesting patterns.
Concerning the limit, we can find that 
$$\displaystyle \lim_{n\rightarrow \infty} S^{(j)}_n=(-1)^j ((j-1)!-!(j-1)\,e)$$
May be, the asymptotics could be of interest
$$S^{(j)}_n=(-1)^j((j-1)!-!(j-1)\,e) +(-1)^{j+1}\frac{(j+1)!-!(j+1)\,e }{2n}+O\left(\frac{1}{n^2}\right)$$
For $j=3$ and $n=50$, the exact value is $\approx 0.71370532$ while the asymptotics leads to  $\approx 0.71363646$.
Update
Following this question of mine, the asymptotics write $$S^{(j)}_n=(-1)^j\left(\left(\alpha_0-\beta_0e\right)-\frac{\left(\alpha_1-\beta_1e\right)}{2n}+\frac{\left(\alpha_2-\beta_2e\right)}{24n^2}\right)+O\left(\frac{1}{n^3}\right)$$ with $$\alpha_0=(j-1)!\qquad \qquad \beta_0=!(j-1)$$ $$\alpha_1=(j+1)!\qquad \qquad \beta_1=!(j+1)$$ $$\alpha_2= 3\times(j+3)! - 8\times(j+2)! \qquad \qquad \beta_2=3\,\times\,!(j+3) - 8\,\times\,!(j+2)$$ Many thanks to achille hui who identified the sequence for $\beta_2$ and provided a nicer expression for $\alpha_2$.
For $j=3$ and $n=50$, the exact value is $\approx 0.71370532$ while the new asymptotics leads to  $\approx 0.71370644$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\color{#f00}{\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} \over n^{k}}\int_{0}^{1}x^{k + 2}\,\dd x =
\lim_{n \to \infty}
\int_{0}^{1}x^{2}\sum_{k = 0}^{n}{n \choose k}\pars{x \over n}^{k}\,\dd x
\\[3mm] = &\
\lim_{n \to \infty}\
\underbrace{\int_{0}^{1}x^{2}\pars{1 + {x \over n}}^{n}\,\dd x}
_{\ds{\color{red}{\large\S}}}\ =\
\int_{0}^{1}x^{2}\expo{x}\,\dd x -
\lim_{n \to \infty}\int_{0}^{1}x^{2}
\bracks{\expo{x} - \pars{1 + {x \over n}}^{n}}\,\dd x
\\[5mm] = &\
\color{#f00}{\expo{} - 2} -
\lim_{n \to \infty}\int_{0}^{1}x^{2}
\bracks{\expo{x} - \pars{1 + {x \over n}}^{n}}\,\dd x
\end{align}
Indeed, the $\ds{\color{red}{\large\S}}$-integral has a closed expression. Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}x^{2}\pars{1 + {x \over n}}^{n}\,\dd x} =
{1 \over n^{n}}\int_{n}^{n + 1}\pars{x - n}^{2}x^{n}\,\dd x
\\[5mm] = &\
{1 \over n^{n}}\bracks{%
\int_{n}^{n + 1}x^{n + 2}\,\dd x -
2n\int_{n}^{n + 1}x^{n + 1}\,\dd x +
n^{2}\int_{n}^{n + 1}x^{n}\,\dd x
}
\end{align}
A: Assuming that $\binom{n}{k}=0$ for $k>n$, we have $\frac{\binom{n}{k}}{n^k(k+3)}\le\frac1{(k+3)k!}$ for all $k,n\ge0$. Therefore, we can apply Dominated Convergence. If we note that for each $k$, the terms increase in $n$, we can also apply Monotone Convergence.
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^k(k+3)}
&=\lim_{n\to\infty}\sum_{k=0}^n\frac{n(n-1)\cdots(n-k+1)}{n^k(k+3)k!}\\
&=\sum_{k=0}^\infty\frac1{(k+3)k!}\\
&=\sum_{k=0}^\infty\frac{(k+1)(k+2)}{(k+3)!}\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+3)-2(k+3)+2}{(k+3)!}\\
&=\underbrace{\ \sum_{k=1}^\infty\frac1{k!}\ }_{e-1}\underbrace{-2\sum_{k=2}^\infty\frac1{k!}+2\sum_{k=3}^\infty\frac1{k!}}_{-1}\\[3pt]
&=e-2
\end{align}
$$
A: Put $\displaystyle u_{n,k}=\frac{(n-1)\cdots (n-k+1)}{n^{k-1}}$ for $1\leq k\leq n$, and $u_{n,0}=1$, $u_{n,k}=0$ for $k\geq n+1$,we have 
$$S_n=\sum_{k=0}^n \frac{(n-1)\cdots (n-k+1)}{n^{k-1}k!(k+3)}=\sum_{k\geq 0}\frac{u_{n,k}}{k!(k+3)}$$
We have $0\leq u_{n,k}\leq 1$ for all $n,k$, and for fixed $k$, $u_{n,k}\to 1$ if $n\to +\infty$. Let $\mu$ be the measure on $\mathbb{N}$ such that $\mu(\{n\})=1$ for all $n$. Then the function $v_n$ defined by $\displaystyle v_n(k)=\frac{u_{n,k}}{k!(k+3)}$ is in $L^1(\mu)$ for all $n$, has for (simple) limit the sequence $v$ defined by $\displaystyle v(k)=\frac{1}{k! (k+3)}$, and is bounded by $w=v\in L^1(\mu)$ independant of $n$. Hence by the Dominated Convergence Theorem, we get that $\displaystyle S_n\to \sum_{k\geq 0}\frac{1}{k!(k+3)}$, that is easy to compute.  
