# Evaluate : $\lim_{n\to \infty} \sum_{r=0}^{n} \frac{\binom nr}{(r+4)n^r}$

Question
Evaluate : $$\lim_{n\to \infty} \sum_{r=0}^{n} \frac{\binom nr}{(r+4)n^r}$$

I have come across a similar question: Evalute $\lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)}$ . The problem is that the linked question asks for an approach that doesn't require the use of Riemann sums (or definite integrals). However, I'm looking for an approach that does involve definite integration.

I do know that I should try manipulating the sum in such a way that I can get a term of $$\frac{r}{n}$$ and $$\frac{1}{n}$$ but I can't figure out a way to do so.

• Did you see Felix's answer in the post you attached? Does it qualify? – Teresa Lisbon Sep 14 '20 at 6:59
• @TeresaLisbon yes I have, but I'm not sure I really understood it. – sai-kartik Sep 14 '20 at 7:02

$$\frac1{r+a}=\int_0^1 x^{r+a-1}dx$$
where $$a$$ is a constant equal to $$3$$ in the case of that question, then interchanging the order of integration and summation (this is valid due to uniform convergence). Then, the binomial theorem is used to note that $$\sum_k\binom{n}{k}\left(\frac{x}{n}\right)^k=\left(1+\frac{x}{n}\right)^n.$$ At the end, the limit is brought into the integral to turn $$\lim(1+x/n)^n$$ into $$e^x$$. The rest is just evaluating an elementary integral (you can do this, for instance, by integration by parts).
$$L=\lim_{n \to \infty} \int_{0}^{1}\sum_{r=0}^{n} {n \choose r}\frac{ t^{r+3}}{n^r} dt= \int_{0}^{1} t^3 \lim_{n \to \infty}(1+t/n)^n dt= \int_{0}^{1} t^3 e^{t} dt$$ $$\implies L=[e^{t}(t^3-3t^2+6t-6)]_{0}^{1}=6-2e$$