Asked by Sean
limit of (10^n)/(n!) as n->infinity
The answer is zero. Can you prove it?
The answer is zero. Can you prove it?
Answers
Answered by
Count Iblis
What you need to do is prove that for every epsilon > 0, you can find an N such that for all n > N you have that
10^n/n! < epsilon
You can write
10^n/n! = Product from k=1 to n of 10/k
In this product the first ten factors are larger than 1. For k > 10 the factors are smaler than 1. If we call the product of the first ten factors F, then for n > 10, we clearly have:
10^n/n! < F*10/n
because if we omit all te factors between k = 11 and k = n-1, we are clearly overestimating the product.
This means that if you take N such that
F*10/N < epsilon -------->
N > 10 F/epsilon
then we can be sure that for all n > N the product is less than epsilon.
10^n/n! < epsilon
You can write
10^n/n! = Product from k=1 to n of 10/k
In this product the first ten factors are larger than 1. For k > 10 the factors are smaler than 1. If we call the product of the first ten factors F, then for n > 10, we clearly have:
10^n/n! < F*10/n
because if we omit all te factors between k = 11 and k = n-1, we are clearly overestimating the product.
This means that if you take N such that
F*10/N < epsilon -------->
N > 10 F/epsilon
then we can be sure that for all n > N the product is less than epsilon.
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