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How can I prove that the result of the following limit is rational/irrational?$$ \lim_{n\to \infty}\left(\sum^{n}_{r=0} \frac{\binom{n}{r}}{n^{r}(r+3)}\right)$$

Would solving this limit satisfy? How would I solve this? So, if the result came out to be say $\pi$, can we conclude that it is irrational? Meaning, is it a complete proof?

Caleb Stanford
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trunkst
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  • If you find an expression for the limit, as Yagna Patel did,you may be able to see (as here) whether it is rational or not. Here we can tell. If the limit were (for example) $e^2+\sqrt 2 + \pi$ we still wouldn't know. We would "know" it was irrational (because most numbers are), but wouldn't have a proof. – Ross Millikan Aug 15 '15 at 05:10
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    How about changing 3 to a, letting a vary and then ask about the values of a that would make the limit rational or irrational? – BCLC Aug 15 '15 at 05:18
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    @BCLC. If $\text{Re}(a)>0$, then $\displaystyle F(a):=\lim_{n\to\infty},\sum_{r=0}^n,\frac{\binom{n}{r}}{n^r,(r+a)}=\int_0^1 ,x^{a-1},\exp(x),\text{d}x=(-1)^{-a},\gamma(a,-1)$, where $\gamma$ is the lower incomplete gamma function. I don't think we can hope to determine all $a$'s such that $F(a)$ is rational. What is certainly the case is that $F(a)$ is irrational for all $a\in\mathbb{N}$. – Batominovski Aug 15 '15 at 06:08

2 Answers2

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Let $f_n(x)=\sum_{r=0}^n\,\binom{n}{r}\,\frac{x^r}{r+3}$ for $x\in\mathbb{R}$ and $n\in\mathbb{N}$. Then, $$\frac{\text{d}}{\text{d}x}\,\left(x^3\,f_n(x)\right)=\sum_{r=0}^n\,\binom{n}{r}\,x^{r+2}=x^2(1+x)^n\,.$$ Because $x^3f_n(x)$ is $0$ when $x=0$, we conclude that $$f_n(x)=x^{-3}\,\int_0^x\,t^2(1+t)^n\,\text{d}t\,.$$ Write $a_n:=f_n\left(\frac{1}{n}\right)$. We are looking for $\displaystyle\lim_{n\to\infty}\,a_n$.

Note that $$a_n=n^3\,\int_0^{1/n}\,t^2(1+t)^n\,\text{d}t=\int_0^1\,s^2\left(1+\frac{s}{n}\right)^n\,\text{d}s\,,$$ where $s:=nt$. Let $g_n(s):=s^2\left(1+\frac{s}{n}\right)^n$ for $n\in\mathbb{N}$ and $s\in[0,1]$. Observe that $g_n$ converges uniformly (or increasingly) to $g$ as $n\to\infty$, where $g(s):=s^2\exp(s)$ for all $s\in[0,1]$. Hence, we can switch the limit and the integral, and obtain $$\lim_{n\to\infty}\,a_n=\int_0^1\,s^2\exp(s)\,\text{d}s=\text{e}-2\,,$$ which is irrational.

In general, if $F_n(a,x):=\sum_{r=0}^n\,\binom{n}{r}\,\frac{x^r}{n^r(r+a)}$ for $n\in\mathbb{N}$, $a\in\mathbb{C}$ with $\text{Re}(a)>0$, and $x\in\mathbb{C}$, then $$F_n(a,x)=\int_0^x\,s^{a-1}\,\left(1+\frac{s}{n}\right)^n\,\text{d}s\,.$$ Hence, $$\lim_{n\to\infty}\,F_n(a,x)=\int_0^x\,s^{a-1}\,\exp(s)\,\text{d}s=(-1)^{-a}\,\gamma(a,-x)\,,$$ where $\gamma$ is the lower incomplete gamma function.

Batominovski
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$$\begin{equation} \begin{split} \lim_{n \to \infty} { \sum_{r=0}^n {\frac{\left( _r^n \right)}{n^r(r+3)}}} & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{n!}{r!(n-r)!n^r(r+3)}}} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\left[ {\lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right)} \left( \frac{1}{r!(r+3)} \right)\right]}} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{1}{r!(r+3)}}} \quad \quad \color{green} {\text{Note: } \lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right) = 1} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{{x^{r+3}}}{r!(r+3)}}} \quad \quad \color{green}{\text{Note: For } x = 1} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n { \int_0^1 {\frac{x^{r+2}}{r!}dx}}} \\ & = \int_0^1 x^2 \lim_{n \to \infty} \sum_{r=0}^n {\frac{x^r}{r!}} dx \\ & = \int_0^1 {x^2e^x \space dx} \\ & = \left[x^2e^x\right]_0^1 - 2 \int_0^1 {xe^x \space dx} \\ & = e - 2 [e-(e-1)] \\ & = e - 2 \end{split} \end{equation} $$

$\therefore \text{We can successfully conclude that} \displaystyle\lim_{n \to \infty} { \sum_{r=0}^n {\frac{\left( _r^n \right)}{n^r(r+3)}}} \text{ is irrational.} $

  • Thank You for your help. This made it crystal clear – trunkst Aug 15 '15 at 02:14
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    I didn't downvote, but I imagine the downvote has something to do with the nature of this answer as a series of equations without offering intuition about how the equations were arrived at. – Milo Brandt Aug 15 '15 at 02:24
  • @MiloBrandt I see. I'll try to update my solution. –  Aug 15 '15 at 02:27
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    I'm the second downvote. This is very sloppy. The second and third equalities are maybe true, but they are certainly not obvious and definitely not justified by the fact that part of the term of the series converges to 1. The interchange between the series and the integral a few lines later is less suspect, but still should be justified. – Martin Argerami Aug 15 '15 at 05:44