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annoymage
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Homework Statement
ln(n) < n1/4 for n > N0
i missed this lecture, so i can't understand what N0 exactly mean
can anyone tell me what it is?
Better stated,annoymage said:Homework Statement
ln(n) < n1/4 for n > N0
i missed this lecture, so i can't understand what N0 exactly mean
can anyone tell me what it is?
ln n < n1/4 do satisfy for all integer n >0
You mistakenly assumed N0 is zero. It's not.annoymage said:waaaaaaaa, so its wrong then, where did i make mistake?
annoymage said:ln n < n1/4 do satisfy for all integer n > 0
so as
ln n < n1/k , k>0 do also satisfy for all integer n > 0.
because,
lim (n->infinity) [tex]\frac{ln n}{n^(1/k)}[/tex] for k>0
= lim (n->infinity) [tex]\frac{k}{n^1/k}[/tex] (using Lhospital rule)
= 0 < 1
=> [tex]\frac{ln n}{n^(1/k)}[/tex] < 1 for all k>0
and it satisfy for all values of n>0
but if the above state that n > N0
its not integer start with 0 right? because n=0 is undefine
annoymage said:im actually finding whether [tex]\sum[/tex] [tex]\frac{(ln n)^3}{n^3}[/tex] converges or diverges.. and I am using comparison test
ln n < n1/4 , n>N0
(ln n)3 < n3/4
=> [tex]\frac{(ln n)^3}{n^3}[/tex] < [tex]\frac{1}{n^(9/4)}[/tex]
since [tex]\sum[/tex][tex]\frac{1}{n^(9/4)}[/tex] converges
by comparison test [tex]\sum[/tex][tex]\frac{(ln n)^3}{n^3}[/tex] also converges
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annoymage said:ln n < n1/4 do satisfy for all integer n > 0
so as
ln n < n1/k , k>0 do also satisfy for all integer n > 0.
because,
lim (n->infinity) [tex]\frac{ln n}{n^(1/k)}[/tex] for k>0
= lim (n->infinity) [tex]\frac{k}{n^1/k}[/tex] (using Lhospital rule)
= 0 < 1
=> [tex]\frac{ln n}{n^(1/k)}[/tex] < 1 for all k>0
and it satisfy for all values of n>0
but if the above state that n > N0
its not integer start with 0 right? because n=0 is undefine
annoymage said:im actually finding whether [tex]\sum[/tex] [tex]\frac{(ln n)^3}{n^3}[/tex] converges or diverges.. and I am using comparison test
ln n < n1/4 , n>N0
(ln n)3 < n3/4
=> [tex]\frac{(ln n)^3}{n^3}[/tex] < [tex]\frac{1}{n^(9/4)}[/tex]
since [tex]\sum[/tex][tex]\frac{1}{n^(9/4)}[/tex] converges
by comparison test [tex]\sum[/tex][tex]\frac{(ln n)^3}{n^3}[/tex] also converges
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Have you read my posts, annoymage?annoymage said:(regarding ln n < n1/4 , n>N0)
i really believe this is what i saw on lecture.
maybe N0 is a different thing, is it?
D H said:Better stated,
[tex]\exists N\in \mathbb N \,:\,\ln n < n^{1/4}\,\forall\,n>N_0[/tex]
In English, there is some positive integer N0 such that [itex]\ln n < n^{1/4}[/itex] for all n>N0.
In other words N0 is some integer; you probably need to find it.
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