monotonicity of [tex]\frac{1}{(logn)^n}[/tex]
- Thread starter mathmari
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Hi!!!How can I find the monotonicity of \(\displaystyle \frac{1}{(logn)^n}\) ?
what is the chief property of monotonic functions? What does this mean about their derivative?
I want to show the monotonicity by using sequences and not functions..How can I do this in this way??
it's virtually the same problem, but if you like you can attack it another way.
let S[n] = (log(n))-n
what can you say about the ratio of S[n+1]/S[n]? It might help to ask whether log(n) is greater or less than log(n+1) first, and if log(n+1) is greater than 0.
n+1>n
Because of the fact that logn is increasing=> log(n+1)>log(n)
and also, n+1>1 sine logn is increasing,we have log(n+1)>log(1)=0
And how can I conclude something about the monotonicity?
sigh...
S[n+1]/S[n] = log(n+1)n+1 / log(n)n
= (log(n+1)/log(n))n * log(n+1)
now you showed above that log(n+1) > log(n), so log(n+1)/log(n) > 1 and thus the nth power of that is greater than 1.
you further showed that for n>1 log(n+1) > 0 so the product is also greater than 1.
So the ratio of S[n+1]/S[n] is always greater than 1.
What does that tell you about whether S is always increasing, i.e. is monotonically increasing?
Oh yes...It tells me that S is always increasing.. And how can I find the limit of the sequence??
Think about it for a bit. You have the reciprocal of an ever increasing number... what do you think the limit will be?
No. Have you even tried plotting the sequence to see what it looks like? Do this for the first 10 or so terms and then determine what the limit is again.