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\begin{align*} & \lim\limits_{j \to \infty}{j^{\,j/2} \over j!} \\ \end{align*}

This problem is from a real analysis textbook in the chapter on the natural log and properties of exponents. I'm struggling with how to approach this. I don't think you would use L'Hopital's rule.

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  • $\begingroup$ Take logarithms. $\endgroup$ – Rafa Budría Jan 1 at 22:21
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    $\begingroup$ Ratio test plus the fact that $(1+1/n)^n\to e$ should do it for you (the ratio of successive terms goes to zero). $\endgroup$ – anthonyquas Jan 1 at 22:22
  • $\begingroup$ Do you know about Stirling's approximation? $\endgroup$ – Tom Himler Jan 1 at 22:22
  • $\begingroup$ What book is this? $\endgroup$ – Symposium Jan 1 at 22:44
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    $\begingroup$ Krantz Real Analysis 4th edition. ISBN 978-1498777681. Page 225. $\endgroup$ – clay Jan 2 at 17:17
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Let $L$ be the limit. Then, it is :

$$L = \lim_{n \to \infty} \frac{n^{n/2}}{n!} \Leftrightarrow \ln L = \lim_{n \to \infty} \left(\ln n^{n/2}-\ln n! \right)$$ $$\Leftrightarrow$$ $$\ln L =\lim_{n \to \infty} \left(\frac{n}{2}\ln n - \ln n!\right)$$

Now we will use the fact

$$\ln n! = n \ln n - n + \mathcal{O}\left(\ln n\right)$$

which is called Stirling's Approximation (actually it is a consquence of the original formula, more information can be found in the link and credits to Jack D'Aurizio for mentioning it as well).

$$\begin{align*}\ln L &=\lim_{n \to \infty} \left(\frac{n}{2}\ln n - n \ln n + n - \mathcal{O}\left(\ln n\right)\right)\\&=\lim_{n \to \infty} \left(n-\frac{n}{2}\ln n\; -\mathcal{O}\left(\ln n\right)\right) \end{align*}$$

But $\mathcal{O}\left(\ln n\right) \to 0$ and you should be able to finish now.

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    $\begingroup$ Such approximation is just a consequence of the rectangle or trapezoid rule, the actual Stirling's approximation is $$ n! = \left(\frac{n}{e}\right)^n \sqrt{2\pi n}\left(1+O\left(\frac{1}{n}\right)\right)$$ incorporating the obscure constant $\Gamma\left(\frac{1}{2}\right)$. $\endgroup$ – Jack D'Aurizio Jan 1 at 22:28
  • $\begingroup$ @JackD'Aurizio True point ! Thanks a lot for adding it as well. $\endgroup$ – Rebellos Jan 1 at 22:29
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The ratio text is a more elementary tool: $$\frac{(j+1)^\tfrac{j+1}2}{(j+1)!}\,\frac{j!}{j^{\tfrac j2}}=\frac{(j+1)^{\tfrac{j}2}\sqrt{j+1}}{(j+1)j^{\tfrac j2}}=\underbrace{\sqrt{\Bigl(1+\frac1j\Bigr)^j}}_{\begin{array}{c}\downarrow\\\sqrt{\mathrm e}\end{array}}\,\frac1{\sqrt{j+1}}\to 0$$

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  • $\begingroup$ The ratio test shows that that the corresponding infinite series will in fact converge, and since it converges the terms must approach zero. This is the simplest answer. Thanks! $\endgroup$ – clay Jan 2 at 18:19
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Without invoking Stirling's approximation, the Hermite-Hadamard inequality, the $\log$-convexity of $\Gamma$ or the trapezoid rule (all viable approaches), one may simply notice that by defining $$ a_n = \frac{n^n}{n!^2} $$ we have $$ \frac{a_{n+1}}{a_n} = \frac{1}{n+1}\underbrace{\left(1+\frac{1}{n}\right)^n}_{\text{bounded}} \to 0 $$ as $n\to +\infty$, hence your limit is zero as well.

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