# Defining $a_n=\frac mn$ for $m$ such that $\frac{(m-1)!}{n^{m-1}}\le 1 \lt \frac{m!}{n^m},$ $\lim_{n\to \infty}a_n=e$?

I enjoy making and solving new questions. I made the following question:

Let $n$ be a positive integer. Let's consider the fractions formed as $\frac{a}{n}$ where $a$ is a positive integer. Starting multiplying as $$\frac1n\times\frac2n\times\frac3n\times\cdots,$$then stop it when the product is first over $1$. Letting $a_n$ be the last fraction you multiply, then find $\lim_{n\to \infty}a_n$.

I'm going to write another form of the same question.

Defining $a_n=\frac mn$ for $m$ such that $$\frac{(m-1)!}{n^{m-1}}\le 1 \lt \frac{m!}{n^m},$$ then find $\lim_{n\to \infty}a_n$.

After my observation, I reached the following expectation:

My expectation: $\lim_{n\to \infty}a_n=e.$

I'm very interested in this value, but I'm facing difficulty. Then, here is my question.

Question: Could you show me how to solve the above question?

• Is it possible that you mean to take the limit as $m\to \infty$? Commented Sep 3, 2013 at 15:41
• @abiessu: I don't get what you mean. Our limit is $n\to \infty$. I think $m$ depends on $n$. Commented Sep 3, 2013 at 15:47
• @abiessu I think its the same. Or we could substitute every occurrence of $m$ for $na_n$. Commented Sep 3, 2013 at 15:52
• @mathlove I think this can be quite easily solved with these en.wikipedia.org/wiki/Gamma_function#Alternative_definitions Commented Sep 3, 2013 at 15:54
• @namehere: Thank you. I'll read it. Commented Sep 3, 2013 at 15:59

I've just been able to prove $$\lim_{n\to\infty} a_n=e$$ by an easy way.
Let $a_n=\frac mn$. Here note that $m$ depends on $n$. Then, by the definition of $a_n$, we get \begin{align}\frac 1n\cdot\frac 2n\cdot\cdots\cdot\frac{m}{n}\gt 1\qquad(1)\end{align} \begin{align}\frac 1n\cdot\frac 2n\cdot\cdots\cdot\frac{m-1}{n}\le 1\qquad(2)\end{align} $(1)$ gives us $$1\cdot 2\cdot \cdots\cdot m\gt n^m.$$ This is equal to \begin{align}\sum_{k=1}^m\log k\gt m\log n\qquad(3)\end{align} Now, noting that $y=\log x$ monotonically increases for $x\gt 0$, we get \begin{align}\sum_{k=1}^m\log k\le\int_{1}^{m+1}\log x dx=[x\log x-x]_{1}^{m+1}=(m+1)\log (m+1)-m\qquad(4)\end{align} $(3)(4)$ gives us $$m\log n\lt (m+1)\log (m+1)-m$$$$\iff m-\log(m+1)\lt m\log (m+1)-m\log n=m\log \frac{m+1}{n}=m\left\{\log\frac mn+\log \left(1+\frac 1m\right)\right\}$$ Since $m\gt0$, dividing the both sides by $m$ gives us \begin{align}1-\frac{\log (m+1)}{m}-\log \left(1+\frac 1m\right)\lt\log\frac mn\qquad(5)\end{align} On the other hand, $(2)$ gives us $$1\cdot 2\cdot\cdots\cdot (m-1)\le n^{m-1}.$$ This is equal to \begin{align}\sum_{k=1}^{m-1}\log k\le (m-1)\log n\qquad(6)\end{align} Hence, we get \begin{align}\sum_{k=1}^{m-1}\log k\ge\int_{1}^{m-1}\log x dx=[x\log x-x]_{1}^{m-1}=(m-1)\log (m-1)-m+2\qquad(7)\end{align} $(6)(7)$ gives us $$(m-1)\log(m-1)-m+2\le (m-1)\log n$$ $$\iff (m-1)\log\frac{m-1}{n}\le m-2$$ $$\iff (m-1)\left\{\log\frac mn+\log\left(1-\frac 1m\right)\right\}\le m-2$$ Since $m\gt 1$, dividing the both sides by $m-1$ gives us \begin{align}\log\frac mn\le\frac{m-2}{m-1}-\log\left(1-\frac 1m\right)\qquad(8)\end{align} $(5)(8)$ gives us \begin{align}1-\frac{\log (m+1)}{m}-\log\left(1+\frac 1m\right)\lt\log\frac mn\le\frac{m-2}{m-1}-\log\left(1-\frac 1m\right)\qquad(9)\end{align} Here, since $\frac 1n\cdot\frac 2n\cdot\cdots\cdot\frac{n-1}{n}\lt 1$, we know $m\ge n$. Hence, $m\to\infty$ when $n\to\infty$.
Hence, when $n\to\infty$, we know $$1-\log\frac{\log (m+1)}{m}-\log\left(1+\frac 1m\right)\to 1,$$ $$\frac{m-2}{m-1}-\log\left(1-\frac 1m\right)\to 1.$$ Note that here we use $$\lim_{m\to\infty}\frac{\log (m+1)}{m}=0.$$ Hence, $(9)$ gives us $\lim_{n\to\infty}\log\frac mn=1$. This leads $\lim_{n\to\infty} a_n=\lim_{n\to\infty}\frac mn=e.$ We now know that the proof is completed.