Power Series

[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.

[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.

[roam]

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.

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}}?

[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.

[roam]

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

Roam

[Mark44]

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

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

[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?

[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.