What is the definition of N0 in proving ln(n) < n^1/4 for n > N0?

[annoymage]

Homework Statement

ln(n) < n^1/4 for n > N₀

i missed this lecture, so i can't understand what N₀ exactly mean

can anyone tell me what it is?

Homework Equations

The Attempt at a Solution

 

[annoymage]

or, is it natural number that start from 0?

 

[Leo34005]

Is that easy for you?

 

[annoymage]

easy? i don't understand what you are trying to say

 

[lanedance]

any context to the statement?

why not see if you can test if/under what constraints the statement is true...

 

[D H]

Homework Statement

ln(n) < n^1/4 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,

\exists N\in \mathbb N \,:\,\ln n &lt; n^{1/4}\,\forall\,n&gt;N_0

In English, there is some positive integer N₀ such that \ln n &lt; n^{1/4} for all n>N₀.

In other words N₀ is some integer; you probably need to find it.

 

[annoymage]

im actually finding whether \sum \frac{(ln n)^3}{n^3} converges or diverges.. and I am using comparison test

ln n < n^1/4

(ln n)³ < n^3/4

=> \frac{(ln n)^3}{n^3} < \frac{1}{n^(9/4)}

since \sum\frac{1}{n^(9/4)} converges

by comparison test \sum\frac{(ln n)^3}{n^3} also converges
-----------------------------------------------------------------------

ln n < n^1/4 do satisfy for all integer n > 0

so as

ln n < n^1/k , k>0 do also satisfy for all integer n > 0.

because,

lim (n->infinity) \frac{ln n}{n^(1/k)} for k>0

= lim (n->infinity) \frac{k}{n^1/k} (using Lhospital rule)

= 0 < 1

=> \frac{ln n}{n^(1/k)} < 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

or did i do any mistakes?

 

[Cyosis]

ln n < n1/4 do satisfy for all integer n >0

Try n=5.

 

[annoymage]

waaaaaaaa, so its wrong then, where did i make mistake?

 

[Cyosis]

My point was to give you a simple self check so that you could see that your starting assumption of ln n<n^(1/4) for all n>0 is wrong. So everything that is based on that is wrong as well. As for the rest it is still unclear what you want to do. You have written down some inequality in your original post, but now it appears that you want to figure out if a series converges or not. You would do well by posting the full problem first, then showing your work. In other words please use the template.

 

[D H]

waaaaaaaa, so its wrong then, where did i make mistake?

You mistakenly assumed N₀ is zero. It's not.

 

[annoymage]

ln n < n^1/4 do satisfy for all integer n > 0

so as

ln n < n^1/k , k>0 do also satisfy for all integer n > 0.

because,

lim (n->infinity) \frac{ln n}{n^(1/k)} for k>0

= lim (n->infinity) \frac{k}{n^1/k} (using Lhospital rule)

= 0 < 1

=> \frac{ln n}{n^(1/k)} < 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

yea, this is all wrong. owho

but this

im actually finding whether \sum \frac{(ln n)^3}{n^3} converges or diverges.. and I am using comparison test

ln n < n^1/4 , n>N₀

(ln n)³ < n^3/4

=> \frac{(ln n)^3}{n^3} < \frac{1}{n^(9/4)}

since \sum\frac{1}{n^(9/4)} converges

by comparison test \sum\frac{(ln n)^3}{n^3} also converges
-----------------------------------------------------------------------

i really believe this is what i saw on lecture.

maybe N₀ is a different thing, is it?

 

[annoymage]

ln n < n^1/4 do satisfy for all integer n > 0

so as

ln n < n^1/k , k>0 do also satisfy for all integer n > 0.

because,

lim (n->infinity) \frac{ln n}{n^(1/k)} for k>0

= lim (n->infinity) \frac{k}{n^1/k} (using Lhospital rule)

= 0 < 1

=> \frac{ln n}{n^(1/k)} < 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

yea, this is all wrong. owho

but this

im actually finding whether \sum \frac{(ln n)^3}{n^3} converges or diverges.. and I am using comparison test

ln n < n^1/4 , n>N₀

(ln n)³ < n^3/4

=> \frac{(ln n)^3}{n^3} < \frac{1}{n^(9/4)}

since \sum\frac{1}{n^(9/4)} converges

by comparison test \sum\frac{(ln n)^3}{n^3} also converges
-----------------------------------------------------------------------

i really believe this is what i saw on lecture.

maybe N₀ is a different thing, is it?

 

[D H]

(regarding ln n < n^1/4 , n>N₀)
i really believe this is what i saw on lecture.

maybe N₀ is a different thing, is it?

Have you read my posts, annoymage?

 

[annoymage]

so what is N₀ for this statement to be true?

ln n < n1/4 , for all n>N₀

or my lecturer did mistakes with this statement?

 

[D H]

No, your lecturer did not make a mistake. As far as the value of N₀, you need to find it.

 

[annoymage]

Better stated,

\exists N\in \mathbb N \,:\,\ln n &lt; n^{1/4}\,\forall\,n&gt;N_0

In English, there is some positive integer N₀ such that \ln n &lt; n^{1/4} for all n>N₀.

In other words N₀ is some integer; you probably need to find it.

yeaa, this one,
and i really need to read the definition to prove that ln n < n^1/4 , n < N thing
k, thankyou :D