- #1

Katrique

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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.

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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