February 26th, 2019, 12:33 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 512 Thanks: 79  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: 6,759 Thanks: 696 
Have you tried mathematical induction?

February 26th, 2019, 02:09 PM  #3 
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706 
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. 
February 26th, 2019, 03:12 PM  #4 
Senior Member Joined: Dec 2015 From: somewhere Posts: 512 Thanks: 79 
By cesarostolz 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 
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706  Looks like squaring the factorial makes a difference after all.

February 27th, 2019, 03:34 AM  #6 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  It's quite easy to see why it would: if $n$ is even, we have $\begin{align*} (n!)^2 &= \underbrace{1 \times 1 \times 2 \times 2 \times \dots \times \left(\frac{n}{2} 1\right) \times \left(\frac{n}{2} 1\right)}_{\geq ~ 1 \times 1 \times 2 \times 2 \times \dots \times 2 \times 2 \\ \qquad= 2^{n4}} \times \underbrace{\frac{n}{2} \times \frac{n}{2} \dots (n1) \times (n1) \times n \times n}_{\geq ~ \frac{n}{2} \times \frac{n}{2} \times \dots \times \frac{n}{2} \times \frac{n}{2} \\ \qquad = \left(\frac{n}{2}\right)^{n+4}} \\ & \geq 2^{n4} \times \left(\frac{n}{2}\right)^{n+4} \\ & = \frac{n^{n+4}}{2^8} \end{align*}$ 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: 512 Thanks: 79 
Is the limit 0 or $\displaystyle \infty$ ? But with theorems I applied it gives 0 . 

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