Finding the Interval of Convergence for a 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!

S_n = (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⁺¹a_n be an alternating series. If the sequance $$\left\langle a_{n}\right\rangle$$ is decreasing then ∑(-1)ⁿ⁺¹a_n 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.