Limit as n->infinity of (n^n)/n

[Potatochip911]

Homework Statement

Evaluate the limit $\lim_{n\to\infty} \dfrac{n^n}{n!}$

Homework Equations

Ratio test: $a_{n+1}*\dfrac{1}{a_n}$

The Attempt at a Solution

I was having trouble evaluating this so I tried to use the ratio test which unfortunately leads to $\lim_{n\to\infty} \dfrac{n}{n+1}$ which just ends up being 1 which does not tell me whether the series is convergent or divergent. Alternatively I tried looking at it like this but I couldn't see where to go from there $$\lim_{n\to\infty} \frac{n}{1} \frac {n}{2} \frac {n}{3} \cdots \frac{n}{n} $$

 

[Orodruin]

You can use either the last expression that you gave to bound the expression from below by a diverging function, or you can take the logarithm of everything and apply Stirling's approximation.

 

[Potatochip911]

You can use either the last expression that you gave to bound the expression from below by a diverging function, or you can take the logarithm of everything and apply Stirling's approximation.

Sorry I'm quite lost as to how I would go about finding a boundary for this function, is there a theorem or something I should be aware of?

 

[Noctisdark]

Hi potatochip911,
n^n/n! = (n*n*...n)/(1*2*...n)
(n+1)^(n+1)/(n+1)! = (n+1)(n+1).../(1*2*...(n+1)),
Hope this makes it easier, you can take out (n+1) from the second and compare the two series, and use induction, if the n^n/n! Is divergent then so is (n+1)^(n+1)/(n+1)!

 

[Orodruin]

Sorry I'm quite lost as to how I would go about finding a boundary for this function, is there a theorem or something I should be aware of?

How about n/k > 1 for n > k?

 

[Potatochip911]

Hi potatochip911,
n^n/n! = (n*n*...n)/(1*2*...n)
(n+1)^(n+1)/(n+1)! = (n+1)(n+1).../(1*2*...(n+1)),
Hope this makes it easier, you can take out (n+1) from the second and compare the two series, and use induction, if the n^n/n! Is divergent then so is (n+1)^(n+1)/(n+1)!

I don't think I'm allowed to use induction to solve these problems.

How about n/k > 1 for n > k?

Can I just do this? $$\frac{n^2}{2} < \frac{n^n}{n!} \mbox{on the interval} [1,\infty] \\ \lim_{n\to\infty} \frac{n^2}{2}=\infty$$ and then since a function smaller function diverges the original must also diverge? Edit: I think I messed this up again I keep thinking in terms of sums

 

[Orodruin]

Can I just do this? $$\frac{n^2}{2} < \frac{n^n}{n!} \mbox{on the interval} [1,\infty] \\ \lim_{n\to\infty} \frac{n^2}{2}=\infty$$ and then since a function smaller function diverges the original must also diverge?

You can even do this: $n = n/1 < n^n/n!$

 

[PeroK]

I don't think I'm allowed to use induction to solve these problems.Can I just do this? $$\frac{n^2}{2} < \frac{n^n}{n!} \mbox{on the interval} [1,\infty] \\ \lim_{n\to\infty} \frac{n^2}{2}=\infty$$ and then since a function smaller function diverges the original must also diverge? Edit: I think I messed this up again I keep thinking in terms of sums

You don't need induction, although it's perfectly valid to use induction for any maths problem - unless it's explicitly stated that you can't. Anyway, back to your problem. Where did you get $n^2/2$ from?

$n^n$ and $n!$ both have $n$ terms; $n^n$ is clearly much larger than $n!$ as $n$ increases. So, all you need is a crude estimate. There is, in fact, a very simple estimate. Can you spot it?

PS: I see Orodruin beat me to it!

PPS: You are also allowed to calculate some values. For example, if you tried n = 5, you'd see that:

$5! = 120$ whereas $5^5 = 25 \times 125$

So, it should be fairly obvious from that where this sequence is heading!

 

[Orodruin]

Where did you get n^2/2 from?

I would say he used the first two factors of:

$$\lim_{n\to\infty} \frac{n}{1} \frac {n}{2} \frac {n}{3} \cdots \frac{n}{n} $$

rather than just the first.

Edit: Of course, this does not tell you how fast it diverges. Stirling's approximation will do that for you (and you should find that it diverges exponentially).

 

[Potatochip911]

You can even do this: $n = n/1 < n^n/n!$

Okay I will keep that in mind for the future.

You don't need induction, although it's perfectly valid to use induction for any maths problem - unless it's explicitly stated that you can't. Anyway, back to your problem. Where did you get $n^2/2$ from?

$n^n$ and $n!$ both have $n$ terms; $n^n$ is clearly much larger than $n!$ as $n$ increases. So, all you need is a crude estimate. There is, in fact, a very simple estimate. Can you spot it?

PS: I see Orodruin beat me to it!

PPS: You are also allowed to calculate some values. For example, if you tried n = 5, you'd see that:

$5! = 120$ whereas $5^5 = 25 \times 125$

So, it should be fairly obvious from that where this sequence is heading!

Yea its just I'm going through a textbook from start to finish and we haven't touched on induction yet so I didn't want to use it. Thanks for the help guys!