Compute limn→∞(n√log|1+(1n⋅log(n))k|).
What I have: ∀ x≥0 : x−x22≤log(1+x)≤x.
Apply to get that the limit equals 1 for any real number k.
Is this correct? Are there any other proofs?
Answer
Yes it works, here's another proof using a little more sofisticate tool (in this case unnecessary, but sometimes more useful).
By Stolz-Cesaro if (xn) is a positive sequence and
limnxn+1xn=l
then
limnn√xn=l.
Taking as (xn) the sequence you defined, an easy calculation shows that xn+1xn→1,
therefore the thesis.
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