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

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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}

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. $$ \begin{align} \lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^k(k+3)} &=\lim_{n\to\infty}\sum_{k=0}^n\frac{n(n-1)\cdots(n-k+1)}{n^k(k+3)k!}\\ &=\sum_{k=0}^\infty\frac1{(k+3)k!}\\ &=\sum_{k=0}^\infty\frac{(k+1)(k+2)}{(k+3)!}\\ &=\sum_{k=0}^\infty\frac{(k+2)(k+3)-2(k+3)+2}{(k+3)!}\\ &=\underbrace{\ \sum_{k=1}^\infty\frac1{k!}\ }_{e-1}\underbrace{-2\sum_{k=2}^\infty\frac1{k!}+2\sum_{k=3}^\infty\frac1{k!}}_{-1}\\[3pt] &=e-2 \end{align} $$

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.

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 $n$)