Limit related Binomial Problem

Question

Evaluate

$\lim_{n\rightarrow∞}\left(\sum\limits_{r=0}^{n}{\left(\frac{n \choose r}{n^r}*\left(\frac{1}{r+3}\right)\right)}\right)$

My approach is as follow but not able to integrate it or convert into a limit series as it involves binomial function $^nC_r$

$\left(\frac{1}{r+3}\right)=\int\limits_{0}^{1}{x^{r+2} dx}\implies \lim\limits_{n\rightarrow∞}{\sum\limits_{r=0}^{n}{\left(\frac{n \choose r}{n^r}*\int\limits_{0}^{1}{x^{r+2} dx}\right)}}=\lim\limits_{n\rightarrow∞}{\sum\limits_{r=0}^{n}{\left({n \choose r}*\int\limits_{0}^{1}{x^2*\left(\frac{x}{n}\right)^r dx}\right)}}$

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$Your approach looks good. Denote by $I_n$ the sum $\sum_{r=1}^n\binom{n}{r}\frac{1}{n^r(r+3)}$, for $n\in\Bbb N$.

Then following your work: $$I_n=\int_0^1x^2\cdot\sum_{r=1}^n\binom{n}{r}(x/n)^r\d x=\int_0^1x^2\cdot[(1+x/n)^n-1]\d x$$By the binomial theorem.

The dominated or monotone convergence theorems apply to say that: $$\lim_{n\to\infty}I_n=\int_0^1\lim_{n\to\infty}x^2\cdot[(1+x/n)^n-1]\d x=\int_0^1(x^2e^x-x^2)\d x$$

This last integral is elementary and has value: $$e-\frac{7}{3}$$

Answer 2

If you want to go beyond the limit for $$S_n=\sum\limits_{r=0}^{n}{\frac{n \choose r}{n^r}\,\frac{1}{r+3}}$$ $$S_n=\frac{n^3 \left(\frac{n+1}{n}\right)^n-2 n^3+2 n^2 \left(\frac{n+1}{n}\right)^n+3 n \left(\frac{n+1}{n}\right)^n+2 \left(\frac{n+1}{n}\right)^n}{(n+1) (n+2) (n+3)}$$ Asymptotically $$S_n=(e-2)-\frac{9 e-24}{2 n}+\frac{443 e-1200}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ is a very good approximation even for small values of $n$ (for $n=10$, the relative error is $0.019$%).