Absolutely convergent/Conditionally convergent/Divergent

[Neon32]

My question is can I use the comparison test? Since Cos is a bounded function?

 

[Mark44]

View attachment 115148

My question is can I use the comparison test? Since Cos is a bounded function?

To use the Comparison Test, you have to have some idea about whether your series converges or diverges. If you believe your series diverges, you have to show that its terms are larger than the corresponding terms of the series you're comparing to. If you believe your series converges, you have to show that its terms are smaller than those of the series you're comparing to.

 

[MAGNIBORO]

If you are trying to check if the sum converges Keep in mind that this sum is not absolutely convergent, because
$$\sum_{n=1}^{\infty}\left |\frac{(-1)^n}{n} \right | = \sum_{n=1}^{\infty}\frac{1}{n}$$

and the harmonic series diverges

I think it would be easier to come up with another method

edit: I made a small reading mistake

 

[Neon32]

To use the Comparison Test, you have to have some idea about whether your series converges or diverges. If you believe your series diverges, you have to show that its terms are larger than the corresponding terms of the series you're comparing to. If you believe your series converges, you have to show that its terms are smaller than those of the series you're comparing to.

I'm a bit confused. Which one do you mean with "Your series"? The new series bn or the orignial series because it matters in this case. If you mean the orignial series.. then how I would know if it converges or not?

 

[Neon32]

According to wikipedia the comparison test is for "infinite series with non-negative (real-valued) terms"
https://en.wikipedia.org/wiki/Direct_comparison_test

If you are trying to check if the sum converges Keep in mind that this sum is not absolutely convergent, because
$$\sum_{n=1}^{\infty}\left |\frac{(-1)^n}{n} \right | = \sum_{n=1}^{\infty}\frac{1}{n}$$

and the harmonic series diverges

This is a different way to solve it and I know. My question was about using the limit comparison test. I'm trying to solve it in different ways

 

[mfb]

You can find convergent series that are larger / smaller for all elements, but that is complicated, and proving the convergence of those is more effort than proving the convergence of your original series.

 

[Neon32]

You can find convergent series that are larger / smaller for all elements, but that is complicated, and proving the convergence of those is more effort than proving the convergence of your original series.

Just to see if I understand it correctly. If an is the original series and bn is another series of my choice. If an<bn and bn is convergent then an is convergent as well
and if an>bn and bn is divergent then an is divergent? This is the theorem

 

[mfb]

If an<bn and bn is convergent then an is convergent as well

That is not sufficient.
b_n=0, a_n=-1
It would be sufficient if a_n>0 for all n, but that is not the case here.

Same problem with the opposite direction.

 

[Mark44]

Which one do you mean with "Your series"?

The series you're working on-- the one in post #1.

My question was about using the limit comparison test.

You said comparison test, not limit comparison test, in post #1. My response was based on what you said.