Prove ln(n)<n(n∈N)
Using only the rule log(xy)=ylog(x)
I tried using any of the known Inequalities am-gm, power mean, titu's lemma, holder's,... but I seem I can't go anywhere near required.
Any hint?
Edit
It was simpler than I thought, it's done with Bernoulli's Inequality.
en>2n=(1+1)n≥1+n>n
ln(en)>ln(n)
ln(n)<n
Answer
Your inequality could be generalized as lnx≤x−1,x>0 To prove it, you perhaps have to use calculus, not just elementary mathematics.Notice that, for f(x)=lnx−x+1,(x>0), f′(x)=1x−1. Thus, f′(x)>0, when $0
But if you have known Bernoulli's inequality, there indeed exists an elementary proof.
Since
(1+x)n≥1+nx,(x≥−1,n∈N+).
Let x=e−1. Then en≥1+n(e−1)=1+ne−n>n.
It follows that n>lnn.
No comments:
Post a Comment