# Thread: Infinite Series, why divergent and not convergent?

1. ## Infinite Series, why divergent and not convergent?

hey guys, I am facing the following problem:

Determine whether the series is convergent or divergent:

$\100dpi&space;\sum_{n=1}^{\infty}&space;\frac{1}{2n}$

The answer in my book says that it is divergent but I do not understand why and how they got to that answer? any help would be greatly appreciated!

thank you!

2. Originally Posted by dmitrip
hey guys, I am facing the following problem:

Determine whether the series is convergent or divergent:

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$

The answer in my book says that it is divergent but I do not understand why and how they got to that answer? any help would be greatly appreciated!

thank you!
because $\displaystyle \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}$ is divergent.

3. thanks for the reply but how do you know that? I thought if I use the test for divergence and take the limit of 1/n, the limit will approach 0 , so wouldn't that mean that it is convergent?

4. Originally Posted by dmitrip
thanks for the reply but how do you know that? I thought if I use the test for divergence and take the limit of 1/n, the limit will approach 0 , so wouldn't that mean that it is convergent?
No. The nth term test says that if the limit of the sequence of terms in the series is not 0, then the series diverges. It says nothing if that limit is 0.

5. Originally Posted by icemanfan
No. The nth term test says that if the limit of the sequence of terms in the series is not 0, then the series diverges. It says nothing if that limit is 0.
ok I understand that but then how do you know that is divergent?

thanks

6. Originally Posted by dmitrip
thanks for the reply but how do you know that? I thought if I use the test for divergence and take the limit of 1/n, the limit will approach 0 , so wouldn't that mean that it is convergent?
The "sequence", which is the list of terms, converges.
The 'series" which is the sum of the terms of the sequence, doesn't.

7. Originally Posted by dmitrip
ok I understand that but then how do you know that is divergent?

thanks
The series $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ is a well-known series called the Harmonic series, and it diverges. Using the fact that any nonzero constant multiple of a divergent series also diverges, we have the result given by skeeter.

8. you need to do some research on the divergence of the harmonic series ...

http://faculty.prairiestate.edu/skif...cs/harmapa.pdf

9. Thanks guys.