Limit of the Sequence n!/n^n

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We find the limit as n approaches infinity of n!/n^n. I hope this helps someone learning about sequences.

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Perfect timing, we were just being taught this. 😅

ericwilliams

This explanation would give you zero points on a calculus exam. "Well the denominator is "clearly" growing faster than numerator so limit is zero", that's mere intuition.

Anyways for those interested the formal proof involves grouping that big fraction like this: (n/n) * (n-1)/n * (n-2) /n * ... * 1/n. We can replace all the terms except for 1/n with just 1 since they are all smaller than 1 since the numerator is smaller than denominator( NOTE: yes 1/n is also smaller than 1 but we didnt replace that with 1 because albeit we would get a correct upper bound of 1 on the original sequence, it wouldn't help us find the limit because we can't employ the squeeze theorem in that case. We got ourselves a bit of a "too large" of an upper bound to be of use; that's why we leave 1/n as is) meaning that the sequence is smaller than 1/n, and we know 1/n has the limit zero so by the squeeze theorem the original sequence converges to 0 as well.

backyard

I've just started self studying these things in calculus, very timely, thanks a lot!! :)

ldisnid

I saw a really awesome solution for this on the Art of Problem solving's calculus textbook. It uses the fact that half of those terms will be less than 1/2. So for 1/n, 2/n, 3/n, ..., through (n/2)/n if n is even or ((n-1)/2 / 2), those terms are all less than or equal to 1/2.

Since the product of all of these terms is less than or equal to 1/2, and the product of all of the other terms are less than or equal to 1, we have the bound n! / n^n is less than or equal to (1/2)^((n)/2).

Hence we have 0 <= n! / n^n <= (1/2)^((n)/2) and as n goes to infinity both sides of this inequality tend toward 0, so by the Squeeze Theorem, the middle function also tends toward 0 as n goes to infinity!

arpanpiano

Please tell me how you made this videos? It will help me alot because I also have a Math channel.

AnuvabSpeaks

I lost track of the number of times on tests that I saved myself by first examining the problem and determining an intuitive answer.
(:

ardiris

That isn't really a proof, 2*n>n for all n but n/(2n) tends to 1/2.

kotazkozla

Broo. Yo estoy vuendi este tema en licenciatura. Por favor poodrias hacer desmostraciones de conveegencia de limites?! Plsss🙏

fernandocupil.

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