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February 26th, 2019, 12:33 PM   #1
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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.
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February 26th, 2019, 01:11 PM   #2
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Have you tried mathematical induction?
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February 26th, 2019, 02:09 PM   #3
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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.
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February 26th, 2019, 03:12 PM   #4
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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.
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February 26th, 2019, 05:01 PM   #5
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Quote:
Originally Posted by idontknow View Post
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.
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February 27th, 2019, 03:34 AM   #6
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Originally Posted by Maschke View Post
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.
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February 27th, 2019, 03:48 AM   #7
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Is the limit 0 or $\displaystyle \infty$ ?
But with theorems I applied it gives 0 .
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