Prove that an=xn/n!→0 for all x
Here is what I tried, but it seems to lead to nowhere.
Choose ϵ>0. We need to show that there exists N∈N such that for all n>N we have |an|<ϵ
So, |(xn/n!)|<ϵ⟹|xn|<n!⋅ϵ (since n! is positive we ignore the absolute signs). So |x|/(ϵ1/n)<[n!(1/n)].
Now I am stuck in solving this for n, and hence finding N ...
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