# Exploring the Radius of Convergence of a Complex Series

• broegger
In summary, the conversation discusses finding the radius of convergence for a given series and proving its convergence and absolute convergence for specific values. The use of the alternating series test and the limit comparison test are also mentioned to aid with proving convergence.
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?

Do you know the alternating series test?

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

Okay...

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

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

Galileo said:
Okay...

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

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

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?

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?

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:

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broegger said:
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:

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?

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

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).

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

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

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## 1. What is the radius of convergence of a complex series?

The radius of convergence of a complex series is a value that determines the range of values for which the series will converge. It is represented by the letter "R" and can be calculated using several methods, including the ratio test and the root test.

## 2. How is the radius of convergence of a complex series calculated?

The radius of convergence can be calculated using the ratio test or the root test. These tests involve taking the limit as n approaches infinity of the ratio or root of the nth term of the series. If the limit is less than 1, the series will converge. If the limit is greater than 1, the series will diverge. If the limit is exactly 1, further tests must be performed to determine the convergence or divergence of the series.

## 3. What is the significance of the radius of convergence?

The radius of convergence is significant because it tells us for which values of the variable in the series, the series will converge or diverge. It also helps us determine the behavior of the series at the endpoints of the interval of convergence.

## 4. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It is always a positive value or infinity.

## 5. What is the interval of convergence of a complex series?

The interval of convergence is the range of values for which the series will converge. It is represented by an interval on the real number line, with the center being the point of convergence and the radius being the radius of convergence. The interval may include or exclude the endpoints, depending on the behavior of the series at those points.

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