When is the test inconclusive for p=1 in UK series?

[foo9008]

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]

 

[Samy_A]

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 $n \in \mathbb N,\ n\neq 0$)?

 

[foo9008]

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 ?

 

[Samy_A]

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.

 

[foo9008]

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 ...

 

[Samy_A]

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.

 

[foo9008]

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 ?

 

[Samy_A]

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
$\displaystyle \sum_{n=1}^N \frac{1}{n²}=\sum_{n=1}^{300} \frac{1}{n²}=\frac{1}{1}+\frac{1}{2²}+\frac{1}{3²}+\dots +\frac{1}{299²}+\frac{1}{300²}$

 

[Mark44]

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.

 

[foo9008]

$N$ 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

 

[Samy_A]

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.

 

[Mark44]

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.