• Support PF! Buy your school textbooks, materials and every day products Here!

Ratio test

  • Thread starter foo9008
  • Start date
  • #1
678
4

Homework Statement


when the test is inconclusive when p = 1? when p=1 , the sum of uk will grow bigger , to infinity , right ?

Homework Equations




The Attempt at a Solution


let's say uk = 2 , uk_2 = 2 , so as uk_3 ,l uk_4 ....... the sum of all of them will beocme infinity , right?[/B]
 

Attachments

Answers and Replies

  • #2
Samy_A
Science Advisor
Homework Helper
1,241
510

Homework Statement


when the test is inconclusive when p = 1? when p=1 , the sum of uk will grow bigger , to infinity , right ?

Homework Equations




The Attempt at a Solution


let's say uk = 2 , uk_2 = 2 , so as uk_3 ,l uk_4 ....... the sum of all of them will beocme infinity , right?[/B]
When p=1, the test is inconclusive, meaning that the series may or may not converge.
Your example indeed diverges.
What about the series defined by ##v_n=\frac{1}{n²}## (for ##n \in \mathbb N,\ n\neq 0##)?
 
  • #3
678
4
When p=1, the test is inconclusive, meaning that the series may or may not converge.
Your example indeed diverges.
What about the series defined by ##v_n=\frac{1}{n²}## (for ##n \in \mathbb N,\ n\neq 0##)?
it will diverge , right ? why it is said to be inconclusive ?
 
Last edited:
  • #4
Samy_A
Science Advisor
Homework Helper
1,241
510
it will converge , right ? why it is said to be inconclusive ?
Yes, it will converge.

This shows that ##\rho =1## doesn't tell you anything about convergence or not of the series.
Your series ##u_n## has a ratio ##\rho =1##, and diverges.
My series ##v_n## has a ratio ##\rho =1##, and converges.

That is what is meant with inconclusive: when ##\rho =1##, you have to find another method to determine whether the series converges or diverges.
 
  • #5
678
4
Yes, it will converge.

This shows that ##\rho =1## doesn't tell you anything about convergence or not of the series.
Your series ##u_n## has a ratio ##\rho =1##, and diverges.
My series ##v_n## has a ratio ##\rho =1##, and converges.

That is what is meant with inconclusive: when ##\rho =1##, you have to find another method to determine whether the series converges or diverges.
sorry , i made a typo in the previous post , i mean it will DIVERGE , let's say we have 1/ (100^2) , we have 200 terms of 1/ (100^2) , we will get sum of them = 0.02 , if there are 300 of them , we will get 0.03 .......
 
  • #6
Samy_A
Science Advisor
Homework Helper
1,241
510
sorry , i made a typo in the previous post , i mean it will DIVERGE , let's say we have 1/ (100^2) , we have 200 terms of 1/ (100^2) , we will get sum of them = 0.02 , if there are 300 of them , we will get 0.03 .......
No, it really converges.

If you take the partial sum up to some number ##N##, you will indeed see that:
##\displaystyle \sum_{n=1}^N \frac{1}{n²} \geq \sum_{n=1}^N \frac{1}{N²} =\frac{N}{N²}=\frac{1}{N}##.
But why would that imply divergence?
(See @Ray Vickson post about the proof that the series converges.)

For the harmonic series, you can group the terms in such a way that it follows that the limit of the partial sums is +∞. (See @stevendaryl 's post.)
You can't do this with this series.
 
  • #7
678
4
No, it really converges.

If you take the partial sum up to some number ##N##, you will indeed see that:
##\displaystyle \sum_{n=1}^N \frac{1}{n²} \geq \sum_{n=1}^N \frac{1}{N²} =\frac{N}{N²}=\frac{1}{N}##.
But why would that imply divergence?
(See @Ray Vickson post about the proof that the series converges.)

For the harmonic series, you can group the terms in such a way that it follows that the limit of the partial sums is +∞. (See @stevendaryl 's post.)
You can't do this with this series.
why the small n is replaced by big N ? they are not the same , right ? the sum of them should be N(1/n^2) , right ?
 
  • #8
Samy_A
Science Advisor
Homework Helper
1,241
510
what do u mean by small n and BIG N ? what's the difference ?
##N## represent a fixed, chosen number, like you chose 200 or 300 in post #5.
##n## is the summation index.

##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{(N-1)²}+\frac{1}{N²}##
If N=300, then
##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\sum_{n=1}^{300} \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{299²}+\frac{1}{300²}##
 
  • #9
33,169
4,854
what do u mean by small n and BIG N ? what's the difference ?
N is a fixed number that is reasonably large. n is the index on the summation, taking on the values 1, 2, 3, ..., up to N.

Also, "text-speak" such as "u" for "you" isn't allowed on this forum.
 
  • #10
678
4
##N## represent a fixed, chosen number, like you chose 200 or 300 in post #5.
##n## is the summation index.

##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{(N-1)²}+\frac{1}{N²}##
If N=300, then
##\displaystyle \sum_{n=1}^N \frac{1}{n²}=\sum_{n=1}^{300} \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{299²}+\frac{1}{300²}##
why the n inside the (1/n^2) not a fixed number? since we have p (ratio) =1
 
  • #11
Samy_A
Science Advisor
Homework Helper
1,241
510
why the n inside the (1/n^2) not a fixed number? since we have p (ratio) =1
Why should it?
The ##n## is not fixed, it is the summation index in ##\displaystyle \sum_{n=1}^{\infty}u_n##.

The ration ##\rho## is defined by ##\rho=\displaystyle \lim_{n \rightarrow \infty}\frac{|u_{n+1}|}{|u_n|}##.
##\rho=1## doesn't mean that ##\frac{|u_{n+1}|}{|u_n|}=1## for some (or all) ##n##, but that the limit of these ratio's is 1.
 
Last edited:
  • #12
33,169
4,854
why the n inside the (1/n^2) not a fixed number? since we have p (ratio) =1
n isn't fixed -- the exponent 2 is fixed. p is not the ratio in the Ratio Test -- it's the exponent in ##\frac 1 {n^p}##.

Please take more time to read these tests more carefully.
 

Related Threads for: Ratio test

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
688
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
824
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
942
  • Last Post
Replies
7
Views
782
  • Last Post
Replies
17
Views
3K
Top