# Math (series)

1. May 9, 2005

### broegger

I am given this series:

$$\sum_{n=1}^\infty\frac{2n}{n^2+1}z^n.$$​

First I have to find the radius of convergence; I find R = 1. Then I have to show that the series is convergent, but not absolutely convergent, for z = -1, i.e. that the series

$$\sum_{n=1}^\infty(-1)^n\frac{2n}{n^2+1}$$​

is convergent, while

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

is not. There is a hint that says $$\frac{2n}{n^2+1}\leq\frac1{n}$$. How can I do this, I know only of the most basic convergence tests?

2. May 9, 2005

### Galileo

Do you know the alternating series test?

3. May 9, 2005

### broegger

No. I guess we're not supposed to use that?

4. May 10, 2005

### Galileo

Okay...

Since they mentioned $$\frac{2n}{n^2+1}\leq\frac1{n}$$, do you know something about the convergence of:

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$?

5. May 10, 2005

### broegger

Yep, converges towards -ln(2). It's obvious how the hint rules out absolute convergence, since $$\sum_{n=1}^{\infty}\frac1{n}$$ is divergent. But the alternating one?

6. May 10, 2005

### Galileo

If you have a sequence $\{a_n\}$ and $a_n \geq 1/n$, then $\sum a_n$ will diverge because the harmonic series does. Here we have $a_n \leq 1/n$ so that comparison fails.

But if you know the alternating harmonic series converges and you combine that with the inequality, can you then show the alternating series to converge?

7. May 10, 2005

### broegger

I'm sorry, I'm an idiot.

The hint says $$\frac{2n}{n^2+1}\geq\frac1{n}$$, that is what's baffling me. Please bear with me :yuck:

Last edited: May 10, 2005
8. May 10, 2005

### cair0

if $$\frac{2n}{n^2+1}$$ is always greater than $$\frac1{n}$$,

and you know that $$\frac1{n}$$ diverges, what do you think that says about something that is larger than it?

9. May 10, 2005

### broegger

It diverges. It's the alternating series I'm having trouble with.

10. May 10, 2005

### shmoe

You might want to look through your notes or textbook again for the alternating series test. Any suggestions I can think of that aren't needlessly awkward essentially just mimic the alternating series test, so you might as well look it up (here it is on Mathworld).

11. May 11, 2005

### scholzie

You may have answered this already, but the alternating series test simply says that for an alternating series which decreases monotonically (or, each term is smaller than the one before it) you have convergence.

The easiest way to use the test is to check if the general term goes to zero. If it does, and does so in a way that $a_{n+1}<a_n$ all the time, then it's convergent. So, since you have the term $$(-1)^n\frac{2n}{n^2+1}$$, which is alternating, and $$\lim_{n\to\infty}\frac{2n}{n^2+1}=0$$, you have convergence.

You can also easily prove that $$\sum_{n=1}^\infty\frac{2n}{n^2+1}$$ Is Divergent by realizing that $$\frac{2n}{n^2+1}$$ acts like $$\frac{1}{n}$$ at infinity, and so by the limit comparison test you have divergence. Of course, you said you already got that one, so that's just me flapping my gums. I don't even know if you've gotten to the LCT yet, but if you hadn't, now you have.

Last edited: May 11, 2005
12. May 11, 2005

### saltydog

Let's try this:
For:
$$\sum_{n=1}^{\infty}(-1)^n a_n$$

If:

$$|a_{n+1}|<|a_n|$$

for all n and:

$$\mathop\lim\limits_{n\to 0}a_n=0$$

The series is convergent as per the Alternating Series Test.

Well, the limit is zero, so simply need to show:

$$\frac{2(n+1)}{(n+1)^2+1}<\frac{2n}{n^2+1}$$

Well, if you subtract:

$$a_{n+1}-a_n$$

You get:

$$\frac{-2(2n^2+n-1)}{(n^2+2n+2)(n^2+1)}$$

That's negative for all n>0. Thus the series converges.

Last edited: May 11, 2005