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

• Potatochip911
In summary, the limit ##\lim_{n\to\infty} \dfrac{n^n}{n!}## can be evaluated by comparing it to the sequence ##n^2/2##, which is much larger than ##n!## as ##n## increases. This sequence diverges exponentially, showing that the original limit also diverges. Calculating values or using Stirling's approximation can further confirm the divergence.
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 $$\lim_{n\to\infty} \frac{n}{1} \frac {n}{2} \frac {n}{3} \cdots \frac{n}{n}$$

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.

Orodruin said:
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?

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)!

Potatochip911 said:
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?

Noctisdark said:
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.

Orodruin said:
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

Potatochip911 said:
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!##

Potatochip911 said:
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!

Last edited:
PeroK said:
Where did you get n^2/2 from?

I would say he used the first two factors of:
Potatochip911 said:
$$\lim_{n\to\infty} \frac{n}{1} \frac {n}{2} \frac {n}{3} \cdots \frac{n}{n}$$
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).

Orodruin said:
You can even do this: ##n = n/1 < n^n/n!##
Okay I will keep that in mind for the future.

PeroK said:
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!

## 1. What does the limit as n approaches infinity of (n^n)/n equal to?

The limit as n approaches infinity of (n^n)/n is equal to infinity. This is because as n gets larger and larger, the numerator (n^n) also gets larger at a much faster rate than the denominator (n). Therefore, the overall value of the fraction approaches infinity.

## 2. Why does the limit as n approaches infinity of (n^n)/n equal to infinity?

This is because as n gets larger and larger, the numerator (n^n) grows at an exponential rate, while the denominator (n) only grows at a linear rate. This results in the overall value of the fraction approaching infinity.

## 3. Can the limit as n approaches infinity of (n^n)/n be evaluated using L'Hopital's rule?

No, L'Hopital's rule cannot be used to evaluate this limit. This is because the denominator (n) is not a constant value, but instead it is also dependent on n. Therefore, L'Hopital's rule cannot be applied in this case.

## 4. Is the limit as n approaches infinity of (n^n)/n a real number?

No, the limit as n approaches infinity of (n^n)/n is not a real number. This is because it is equal to infinity, which is not considered a real number in mathematics.

## 5. Can the limit as n approaches infinity of (n^n)/n be used to find the asymptotic growth rate of a function?

Yes, the limit as n approaches infinity of (n^n)/n can be used to find the asymptotic growth rate of a function. This is because as n gets larger and larger, the overall value of the fraction approaches infinity, which indicates that the function is growing at an exponential rate.

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