# Prove the sequence ((n!)^1/n) / n converges

1. May 27, 2013

### DotKite

1. The problem statement, all variables and given/known data

prove the sequence $\frac{n!^{1/n}}{n}$ converges and find its limit

2. Relevant equations

n/a

3. The attempt at a solution

Ok since this question was in a section having to do with the ratio test, I am making an educated guess that we are suppose to show that

$\sum \frac{n!^{\frac{1}{n}}}{n}$

converges via ratio test, thus implying the sequence $\frac{n!^{1/n}}{n}$ converges to zero

However when I apply the ratio test I end up with,

$\lim_{n\rightarrow\infty} \frac{n(n+1)!^{1/n+1})}{n^{1/n}(n+1)}$

and I am stuck

Last edited: May 27, 2013
2. May 27, 2013

### DotKite

sorry trying to get this latex stuff down. it seems it hasnt worked. one moment

3. May 27, 2013

### Dick

$$\frac{n!^{1/n}}{n}$$ You are really close. Don't put spaces between the 'tex' and the '[' and ']'.

4. May 27, 2013

### DotKite

ok seems to be cleared up!

5. May 27, 2013

### Dick

Good job. You want the limit of the sequence, not the sum of the series. And the ratio test can't help you. If you are allowed to use Stirling's approximation for n!, it's pretty easy. Are you? Otherwise, take the log and think about approximating a sum by an integral.

Last edited: May 27, 2013
6. May 27, 2013

### DotKite

we haven't been taught Stirling's approximation. Nor have we "learned" the approximation via integral. The reason why i am trying to find if the sum converges because if the sum of a_n converges then the sequence {a_n} converges to zero, and that is using all the tools we have been presented in the class

7. May 27, 2013

### Dick

That's a sound strategy, but it won't work. The ratio test will give you 1. Which won't show it converges, which is good because the limit isn't zero. If you don't have something like an integral approximation or Stirling's, I don't see what else to do. Maybe they assigned this problem by mistake?

8. Jun 4, 2013

### klondike

How about using inequality of AM-GM means in conjunction with Sandwich theorem?

9. Jun 4, 2013

### Dick

That's a good idea, it will give you an upper bound for the limit. But it won't give you the value for the limit.

10. Jun 4, 2013

### klondike

Perhaps we can prove the sequence monotonic so that we at least get the converge part. but I don't know how off top of my head.-:(

11. Jun 4, 2013

### Dick

I don't either. It's easy if you've got some asymptotic form for the behavior of n! (like Stirling). n! is pretty hard to deal with if it's got a power like 1/n outside of it.

12. Jun 4, 2013

### johnqwertyful

I was able to do it. Basically, turn n into n^(n/n). Then (n!/n^n)^1/n. Then you can take log on both sides, and use basic properties of logs.

Edit: this might not actually work. It looked a lot neater, but still indeterminate.

13. Jun 4, 2013

### Dick

No, I don't think that will do it.

14. Jun 4, 2013

### johnqwertyful

I don't know if you've covered it, but Cesaro's might come in handy (L'Hopital for sequences).

15. Jun 4, 2013

### Dick

I remain skeptical.

16. Jun 4, 2013

### Office_Shredder

Staff Emeritus
Taking logs is still mildly useful

$$\log\left( \frac{n!^{1/n}}{n} \right) = \frac{1}{n} \log(n!) - \log(n)$$
$$= \frac{1}{n} \left( \log(n) + \log(n-1) + .... + \log(1) \right) - \log(n)$$
$$\leq \log(n) - \log(n) = 0$$
So if $a_n$ is the nth term in the sequence, $\log(a_n) \leq 0$ which shows $a_n \leq 1$ for all n.

If you know what a Riemann sum is you can do way better. The $\leq$ from above came from the inequality
$$\log(n) + \log(n-1) + ... + \log(1) \leq \log(n) + \log(n) +...+ \log(n)$$
Which is obviously true. How bad is this inequality? Well if we move everything to the left hand side we get:
$$\log(n) - \log(n) + \log(n-1) - \log(n) +...+\log(1)-\log(n) = \log(\frac{n}{n}) + \log( \frac{n-1}{n}) + ... + \log(\frac{1}{n}$$
If we include our $\frac{1}{n}$ from way back when this tells us that
$$\frac{1}{n} \left( \log(n) + \log(n-1) + ... + \log(1) \right) - \frac{1}{n} \left(log(n) + \log(n) +...+ \log(n \right) = \frac{1}{n} \left(\log(\frac{n}{n}) + \log( \frac{n-1}{n}) + ... + \log(\frac{1}{n} \right)$$
And this right hand side is a Riemann sum for the integral of log(x) between 0 and 1! So you can figure out what it converges to as n goes to infinity. Going back to the original log form in the first paragraph and using this result gives you what the sequence converges to.

17. Jun 5, 2013

### Dick

I was just replacing the sum of the logs with the integral approximation n*log(n)-n and waving my hands to say the approximation was good enough, but reducing it to a Riemann sum is even better.

18. Jun 5, 2013

### klondike

Dick, my original proposal is a dead end cause even I managed to prove it monotonic, I won't be able to evaluate its limit -- FAILED ATTEMPT. Let me try again.

It's possible to prove $(1+\frac{1}{n})^n$ monotonic increase. As a fact, it converges to e. Hence, $$(1+\frac{1}{n})^{n}<e$$
After some algebraic manipulation, we have
$$\frac{n+1}{ne}<\frac{\sqrt[n]{n!}}{n}$$

It's also possible to prove $(1+\frac{1}{n})^{n+1}$ monotonic decrease. Obviously it converges to e. Hence, $$e<(1+\frac{1}{n})^{n+1}$$
After some similar algebraic manipulation, we have
$$\frac{\sqrt[n]{n!}}{n}<\frac{n+1}{ne}\sqrt[n]{n+1}$$

Combining the two results we now have

$$\frac{n+1}{ne}<\frac{\sqrt[n]{n!}}{n}<\frac{n+1}{ne}\sqrt[n]{n+1}$$

Now we can comfortably evaluate the limits as n approaches infinity and apply sandwich theorem and get the desired result.

19. Jun 5, 2013

### Dick

That would do it. Hate to think how long it would take me to think something like that up though, especially since I still can't figure out the 'algebraic manipulation' steps.

20. Jun 5, 2013

### klondike

$$(1+\frac{1}{n})^{n}<e$$
for n=1
$$\frac{2}{1}<e$$
for n =2
$$\frac{3^2}{2^2}<e$$
for n =3
$$\frac{4^3}{3^3}<e$$
and so on so forth... Multiply them together, we then have
$$\frac{(n+1)^n}{n!}<e^n$$

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