Homework Help: 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.