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

  • Thread starter annoymage
  • Start date
In summary, the statement \ln n < n^{1/4} for n > N0 is true for all positive integers n, where N0 is a positive integer that needs to be found. This statement is important in determining the convergence of the series \sum \frac{(ln n)^3}{n^3} using the comparison test. Further understanding of the definition is needed to prove this statement.
  • #1
annoymage
362
0

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?

Homework Equations


The Attempt at a Solution

 
Physics news on Phys.org
  • #2
or, is it natural number that start from 0?
 
  • #3
Is that easy for you?
 
  • #4
easy? i don't understand what you are trying to say
 
  • #5
any context to the statement?

why not see if you can test if/under what constraints the statement is true...
 
  • #6
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?
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.
 
  • #7
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

(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
-----------------------------------------------------------------------

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

or did i do any mistakes?
 
  • #8
ln n < n1/4 do satisfy for all integer n >0

Try n=5.
 
  • #9
waaaaaaaa, so its wrong then, where did i make mistake?
 
  • #10
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.
 
Last edited:
  • #11
annoymage said:
waaaaaaaa, so its wrong then, where did i make mistake?
You mistakenly assumed N0 is zero. It's not.
 
  • #12
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

yea, this is all wrong. owho

but this

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
-----------------------------------------------------------------------

i really believe this is what i saw on lecture.

maybe N0 is a different thing, is it?
 
  • #13
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

yea, this is all wrong. owho

but this

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
-----------------------------------------------------------------------

i really believe this is what i saw on lecture.

maybe N0 is a different thing, is it?
 
  • #14
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?
Have you read my posts, annoymage?
 
  • #15
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?
 
  • #16
No, your lecturer did not make a mistake. As far as the value of N0, you need to find it.
 
  • #17
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.

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
 

1. What is the scientific method?

The scientific method is a systematic approach to answering questions and solving problems. It involves making observations, forming a hypothesis, conducting experiments, analyzing data, and drawing conclusions.

2. How do scientists conduct experiments?

Scientists conduct experiments by following a set of procedures and controls to test their hypothesis. This may involve manipulating variables, collecting data, and analyzing results to determine if the hypothesis is supported.

3. What is the difference between a theory and a hypothesis?

A hypothesis is a proposed explanation for a phenomenon, while a theory is a well-supported and widely accepted explanation that has been repeatedly tested and confirmed through multiple experiments and observations.

4. How do scientists ensure the accuracy of their results?

Scientists use various methods to ensure the accuracy of their results, such as controlling variables, using large sample sizes, and conducting repeat experiments. They also undergo peer review, where other scientists in the same field review and critique their work.

5. What is the role of statistics in scientific research?

Statistics play a crucial role in scientific research by allowing scientists to analyze and interpret data, identify patterns and relationships, and draw conclusions from their experiments. It also helps to determine the significance and reliability of results.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
3K
Replies
1
Views
836
  • Calculus and Beyond Homework Help
2
Replies
48
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
533
  • Calculus and Beyond Homework Help
Replies
1
Views
452
  • Calculus and Beyond Homework Help
Replies
6
Views
421
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Back
Top