# Thread: Today's calculation of integral #8

1. ## Today's calculation of integral #8

Compute $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$

2. Originally Posted by Krizalid
Compute $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
let $a_n = \frac {n^n}{n!}$ and note that the root test will give the same result as the ratio test. that is, we will use the fact that:

$\lim |a_n|^{1/n} = \lim \left| \frac {a_{n + 1}}{a_n} \right|$

Now compute $\lim \left| \frac {a_{n + 1}}{a_n} \right|$

if i did not make a mistake, the answer should be $e$

3. Yeah, that's the answer, but it is asked a solution which involves integration

4. Originally Posted by Krizalid
Yeah, that's the answer, but it is asked a solution which involves integration
well, you're the Inte-Killer, not me.

I have to go for a while. I'll think about it and maye post a solution with integrals later... maybe

5. Here is a way to do it. But it might be cheating, because it is basically Stirling approximation.

Instead compute the $\ln$ of this sequence,
$\ln \left( \frac{n^n}{n!} \right)^{1/n} = \ln n - \frac{1}{n}\ln (n!)$.

But, $\ln (n!) = \sum_{k=1}^n \ln k \approx \int_1^n \ln x dx = n\ln n - n + 1$ by Euler-Maclaurin formula, the remainder term will happen to go to zero.

Thus, we are left with $1 - \frac{1}{n} \to 1$. Which means $\ln a_n \to 1 \implies a_n \to e$.

6. Originally Posted by Krizalid
Compute $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
$y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n} }$....so $ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b igg)$... $ln(y)=\lim_{n \to {\infty}}\frac{1}{n}ln\bigg(\cdot\frac{n^{n}}{\int _0^{\infty}e^{-x}x^{n}dx}\bigg)$...using l'hopital's rule and much simplification we get $ln(y)=\lim_{n \to {\infty}}\frac{ln(n)n!+n!-\int_0^{\infty}e^{-x}x^n\ln(x)dx}{n!}$...using l'hopitals again gives us $ln(y)=\lim_{n \to {\infty}}\frac{ln(n)e^{-n}n^n+(n-1)!+e^{-n}n^n-ln(n)e^{-n}n^n}{e^{-n}n^{n}}$..simplifying we get $ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}$...now by splitting we get $ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1$...so $ln(y)=1$...so $y=e$

7. Okay, let this be more interesting: no powerful theorems.

As for Mathstud28's solution, no L'Hôpital's Rule of course.

8. How do we know that we can treat this as an integral? Like how did you know to use the ratio test? I was looking at it as just a limit that (I'm pretty sure) goes to infinity.

:/ I don't think I even understand the question.

Edit: or... goes to e :P

9. ## Haha

Originally Posted by Krizalid
Okay, let this be more interesting: no powerful theorems.

As for Mathstud28's solution, no L'Hôpital's Rule of course.
Oh...come on I am only in eleventh grade can't I get soem credit haha...that took me forever haha....thanks for nothing ....haha I am just kidding that was hard for me though....I am happy with myself

10. ## Just wondering

Originally Posted by Mathstud28
$y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n} }$....so $ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b igg)$... $ln(y)=\lim_{n \to {\infty}}\frac{1}{n}ln\bigg(\cdot\frac{n^{n}}{\int _0^{\infty}e^{-x}x^{n}dx}\bigg)$...using l'hopital's rule and much simplification we get $ln(y)=\lim_{n \to {\infty}}\frac{ln(n)n!+n!-\int_0^{\infty}e^{-x}x^n\ln(x)dx}{n!}$...using l'hopitals again gives us $ln(y)=\lim_{n \to {\infty}}\frac{ln(n)e^{-n}n^n+(n-1)!+e^{-n}n^n-ln(n)e^{-n}n^n}{e^{-n}n^{n}}$..simplifying we get $ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}$...now by splitting we get $ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1$...so $ln(y)=1$...so $y=e$
But is this right though?

11. Where Krizalid loves integrals, mathstud28 is fond of l'Hôpital's rule

I would have done like TPH, but i don't see where there is a "powerful theorem" ?

12. Originally Posted by Moo
I would have done like TPH, but i don't see where there is a "powerful theorem" ?
The powerful theorem here is the Euler-Maclurin summation which tells us that we can approximate a sum with an integral with error estimate good enough to vanish in the limit.

13. It appears to me all you need to do is put a bound on $n!$. Of course, Stirling approximation is the best way to go, but that would be cheating here. You can use what PaulRS did here to get good estimates on $n!$ and use squeeze theorem.

14. Originally Posted by Krizalid
Compute $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
$\left\{ {\frac{{n^n }}
{{n!}}} \right\}^{\frac{1}
{n}} = \exp \left\{ {\frac{1}
{n}\ln \frac{{n^n }}
{{n!}}} \right\} = \exp \left\{ {\frac{1}
{n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}
{k}} } \right)} \right\} = \exp \left\{ { - \frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} } \right\}.$

As $n\to\infty,\,{\frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} }$
is a Riemann sum for $\int_0^1\ln x\,dx,$ which is a fairly known integral, it follows that $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}=e.$

15. Originally Posted by Krizalid
$\left\{ {\frac{{n^n }}
{{n!}}} \right\}^{\frac{1}
{n}} = \exp \left\{ {\frac{1}
{n}\ln \frac{{n^n }}
{{n!}}} \right\} = \exp \left\{ {\frac{1}
{n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}
{k}} } \right)} \right\} = \exp \left\{ { - \frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} } \right\}.$

As $n\to\infty,\,{\frac{1}
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
{n}} }$
is a Riemann sum for $\int_0^1\ln x\,dx,$ which is a fairly known integral, it follows that $\lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}=e.$
Cool solution. Is that your own problem?

Page 1 of 2 12 Last