# Is this proof for convergence of 3^n/n! rigorous enough?

Hi,

I am trying to self study analysis and was practicing some problems. I wasn't sure if this solution to one of the problems I came across was rigorous enough.

Basically, by writing down the first few terms of 3^n and n!, I figured I can say 3^n < 3*(n-1)! for all n>=13...without actually calculating the numerical values.

so (3^n/n!)<3/n for n>=13. Therefore, for any ε we can can find N>13 and N>3/ε. And therefore the term 3^n/n! converges to 0. Is this a right solution? Is it rigorous enough? Could you advise me of a better way to prove this?

Thanks.

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Ray Vickson
Homework Helper
Dearly Missed
Hi,

I am trying to self study analysis and was practicing some problems. I wasn't sure if this solution to one of the problems I came across was rigorous enough.

Basically, by writing down the first few terms of 3^n and n!, I figured I can say 3^n < 3*(n-1)! for all n>=13...without actually calculating the numerical values.

so (3^n/n!)<3/n for n>=13. Therefore, for any ε we can can find N>13 and N>3/ε. And therefore the term 3^n/n! converges to 0. Is this a right solution? Is it rigorous enough? Could you advise me of a better way to prove this?

Thanks.
You can prove it by induction. First, you can establish it is true for n = 8. So assuming it is true for n = k >= 8, prove it is also true for n = k+1. (This is EASY.) Thus, it is true for all n ≥ 8.

RGV

I assume, you mean, to prove that 3^n<3(n-1)! for all n>=8. I guess that makes sense..Thanks.