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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.


EDIT: I have also read through this answer on the linked post, but I couldn't understand it. An elaboration of the answer would be helpful to me.

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  • $\begingroup$ Did you see Felix's answer in the post you attached? Does it qualify? $\endgroup$ – Teresa Lisbon Sep 14 '20 at 6:59
  • $\begingroup$ @TeresaLisbon yes I have, but I'm not sure I really understood it. $\endgroup$ – sai-kartik Sep 14 '20 at 7:02
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I will just explain the answer linked in the post, and then hopefully you can adapt that to this case. The main idea in the answer you referenced is to use the trick of writing

$$\frac1{r+a}=\int_0^1 x^{r+a-1}dx$$

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).

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    $\begingroup$ That trick certainly deserves a name, though. $\endgroup$ – sai-kartik Sep 19 '20 at 16:00
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$$L=\lim_{n \to \infty} \int_{0}^{1}\sum_{r=0}^{n} {n \choose r}\frac{ t^{r+3}}{n^r} dt= \int_{0}^{1} t^3 \lim_{n \to \infty}(1+t/n)^n dt= \int_{0}^{1} t^3 e^{t} dt$$ $$\implies L=[e^{t}(t^3-3t^2+6t-6)]_{0}^{1}=6-2e$$

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  • $\begingroup$ This is the same as the idea in the linked answer, without any further elaboration. I doubt the OP can understand this if they couldn't understand the other post. $\endgroup$ – YiFan Sep 15 '20 at 5:38

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