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 February 26th, 2019, 12:33 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92 Limit with sequence How can I solve the limit or which theorem can be used ? $\displaystyle \lim_{n\rightarrow \infty } \frac{1^n +2^n + 3^n +...+n^n }{(n!)^2}$ . Last edited by idontknow; February 26th, 2019 at 12:41 PM. February 26th, 2019, 01:11 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Have you tried mathematical induction? February 26th, 2019, 02:09 PM #3 Senior Member   Joined: Aug 2012 Posts: 2,393 Thanks: 749 My guess off the top of my head is that the limit is $\infty$ because $n^n$ dominates $n!$. In fact $\displaystyle \lim_{n \to \infty} \frac{n!}{n^n} = 0$. Interesting limit from real analysis: lim n!/n^n | John Petrie’s LifeBlag I don't think the additional terms in the numerator are significant, nor is squaring the factorial in the denominator. Again just my impression. Thanks from idontknow February 26th, 2019, 03:12 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92 By cesaro-stolz theorem the limit is $\displaystyle \lim_{n\rightarrow \infty} \frac{n^n }{(n!)^2 }$ . Then by stirling approximation $\displaystyle \lim_{n\rightarrow \infty } \frac{e^{2n} }{n^n }$ . Last edited by idontknow; February 26th, 2019 at 03:17 PM. February 26th, 2019, 05:01 PM   #5
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 Originally Posted by idontknow By cesaro-stolz theorem the limit is $\displaystyle \lim_{n\rightarrow \infty} \frac{n^n }{(n!)^2 }$ . Then by stirling approximation $\displaystyle \lim_{n\rightarrow \infty } \frac{e^{2n} }{n^n }$ .
Looks like squaring the factorial makes a difference after all. February 27th, 2019, 03:34 AM   #6
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 Originally Posted by Maschke Looks like squaring the factorial makes a difference after all.
It's quite easy to see why it would: if $n$ is even, we have

\begin{align*} (n!)^2 &= \underbrace{1 \times 1 \times 2 \times 2 \times \dots \times \left(\frac{n}{2} -1\right) \times \left(\frac{n}{2} -1\right)}_{\geq ~ 1 \times 1 \times 2 \times 2 \times \dots \times 2 \times 2 \\ \qquad= 2^{n-4}} \times \underbrace{\frac{n}{2} \times \frac{n}{2} \dots (n-1) \times (n-1) \times n \times n}_{\geq ~ \frac{n}{2} \times \frac{n}{2} \times \dots \times \frac{n}{2} \times \frac{n}{2} \\ \qquad = \left(\frac{n}{2}\right)^{n+4}} \\ & \geq 2^{n-4} \times \left(\frac{n}{2}\right)^{n+4} \\ & = \frac{n^{n+4}}{2^8} \end{align*}

so $(n!)^2$ grows at least as quickly as $n^n$. A similar argument works when $n$ is odd, but you'd have a bunch of "$\pm \frac{1}{2}$" terms floating about. February 27th, 2019, 03:48 AM #7 Senior Member   Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92 Is the limit 0 or $\displaystyle \infty$ ? But with theorems I applied it gives 0 . Tags limit, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Calculus 5 August 3rd, 2012 03:01 AM nappysnake Calculus 2 December 11th, 2011 12:17 PM everettjsj2 Calculus 5 February 25th, 2010 07:04 PM everettjsj2 Calculus 1 February 22nd, 2010 08:20 PM knowledgegain Real Analysis 3 May 5th, 2009 09:12 AM

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