# Power Series

1. Jan 31, 2009

### roam

Find the interval of convergence of the given series and its behavior at the endpoints:

$$\sum^{+\infty}_{n=1} \frac{(x+1)^n}{\sqrt{n}}$$ $$= (x+1) + \frac{(x+1)^2}{\sqrt(2)}+...$$

The attempt at a solution

Using the ratio test: $$\left|\frac{S_{n+1}}{S_{n}}\right| = \sqrt{\frac{n}{n+1}}\left|x+1\right|$$

Hence, $$lim_{n\rightarrow\infty} \left|\frac{S_{n+1}}{S_{n}}\right|= \left|x+1\right|$$

So, the interval of convergence is |x+1|<1, which implies -1<x<1 which in turn is equal to: -2<x<0

At the right hand endpoint where x = 0 we have the divergent p-series $$\sum^{+\infty}_{n=1} \frac{1}{\sqrt{n}}$$ (with p=1/2).

At the left endpoint x=-2 we get the alternating series $$\sum^{+\infty}_{n=1} \frac{(-1)^n}{\sqrt{n}}$$. The book says this series converges but I used the comparison test & found out that it's NOT!

Sn = (1)n+11/√n diverges by comparison with the p-series ∑1/√n (since p<1)

I’m really confused right now, if it is supposed to converge then why does it diverge by the comparison test? What mistakes did I make?

Thanks.

2. Feb 1, 2009

### NoMoreExams

I'm confused by what you are comparing $$\frac{(-1)^n}{\sqrt{n}}$$ to?

How can you compare it to $$\frac{1}{\sqrt{n}}$$?

The first has negative and positive terms whereas the latter does not.

3. Feb 1, 2009

### roam

So, what else can I compare it to.....?

Why can't we just re-write the first as $$(-1)^{n+1}\frac{1}{\sqrt{n}}$$ and then compare it to the p-series $$\sum \frac{1}{\sqrt{n}}$$?

4. Feb 1, 2009

### NoMoreExams

because $$(-1)^{n+1} \frac{1}{\sqrt{n}}$$ will still have positive and negative terms right? You'd need a test that has "Absolute" in the name :) if anything.

5. Feb 1, 2009

### roam

Hmm, I'm really curious!...
For example, what test can you use for $$\frac{(-1)^n}{\sqrt{n}}$$?

Roam

6. Feb 1, 2009

### Staff: Mentor

How about the alternating series test? You know that one, don't you?

7. Feb 1, 2009

### roam

Oh OK, here's a part of the theorem for alternating series: Let ∑(-1)n+1an be an alternating series. If the sequance $$\left\langle a_{n}\right\rangle$$ is decreasing then ∑(-1)n+1an converges to a sum A.

So, for $$\frac{(-1)^n}{\sqrt{n}}$$, we can write it as $$(-1)^n \frac{1}{\sqrt{n}}$$.

Since $$\left\langle \frac{1}{\sqrt{n}} \right\rangle$$ is a decreasing sequance (& also divergent), the series converge by virtue of the alternating series test.

Is this correct? Does it make any sense now?

8. Feb 1, 2009

### NoMoreExams

Makes sense to me, as an extreme example of why your original logic doesn't work, think about what the following will produce:

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

The fact that you have alternating terms is a pretty big deal.