Question

Show that for all a>0, \(lim_{n\rightarrow \infty}\dfrac {a^n}{n!}=0\)
Please, help

Answer 1

@phi

Answer 2

you can break it up...

\[\frac{ a }{ n } \cdot \frac{ a^{n-1} }{ \left( n-1 \right)! }\]limit of first term is 0 as n -> infinity

Answer 3

You could also try any of your favorite convergence tests

Answer 4

This problem is proving one, not just find out the limit.

Answer 5

My attempt:
Let \(X_n = \dfrac{a^n}{n!}\) \(\forall a>0\) \(X_n>0\) so that By archimedean theorem, \(X_n\) bounded below.

Answer 6

limit of product is product of the limits

Answer 7

Which show lim of \(X_n\) exists
Let try \(X_{n+1}\)

Answer 8

how does it show that the limit exist? it just says it's bounded below. you would have to show that X_n is strictly decreasing.

Answer 9

\(X_{n+1}= \dfrac{a^{n+1}}{(n+1)!}=\dfrac{a}{n+1}*\dfrac{a^n}{n!}=\dfrac{a}{n+1}X_n\)

Answer 10

and it is, after some k...

Answer 11

if n>a, then \(X_n\) decreases
if n<a, then \(X_n\) increases

Answer 12

I think I have to prove both cases, not just find out the lim. Am I right?

Answer 13

Archimedean theorem shows lim X_n exist

Answer 14

you have to break it up to get to the mototone decreasing part. for some n>k>a you'll have
\[\frac{ a^{n-k} }{ n \cdot \left( n-1 \right) \cdots \left( n-k \right)}\cdot \frac{ a^k }{ k! }\]the left one is monotone decreasing because each factor, from \(\frac{a}{n-k}\) to \(\frac{a}{n}\) is less than 1 and it is bounded below by 0.

Answer 15

likewise the term on the right is bounded below and above because it has a finite number of terms, all of which are greater than 0.

Answer 16

i'm not familiar with the archimedian theorem... i'll check it out.

Answer 17

One more question:
What is the logic on:
if X_n is decreasing, and a = lim X_n, then \(lim_{n\rightarrow \infty} (X_{n+1}) = a\)

Answer 18

because X_n has a limit and if it is a sequence, then the difference between successive terms must go to 0.

Answer 19

that is, if X_n -> a => X_n - X_n-1 -> 0 as n -> infinity

Answer 20

Thanks for being patient to me. Please, make it clearer