- #1

roam

- 1,271

- 12

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

[tex]\sum^{+\infty}_{n=1} \frac{(x+1)^n}{\sqrt{n}}[/tex] [tex]= (x+1) + \frac{(x+1)^2}{\sqrt(2)}+...[/tex]

[tex]\sum^{+\infty}_{n=1} \frac{(x+1)^n}{\sqrt{n}}[/tex] [tex]= (x+1) + \frac{(x+1)^2}{\sqrt(2)}+...[/tex]

**The attempt at a solution**

Using the ratio test: [tex]\left|\frac{S_{n+1}}{S_{n}}\right| = \sqrt{\frac{n}{n+1}}\left|x+1\right|[/tex]

Hence, [tex]lim_{n\rightarrow\infty} \left|\frac{S_{n+1}}{S_{n}}\right|= \left|x+1\right|[/tex]

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 [tex]\sum^{+\infty}_{n=1} \frac{1}{\sqrt{n}}[/tex] (with p=1/2).

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

S

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.

Using the ratio test: [tex]\left|\frac{S_{n+1}}{S_{n}}\right| = \sqrt{\frac{n}{n+1}}\left|x+1\right|[/tex]

Hence, [tex]lim_{n\rightarrow\infty} \left|\frac{S_{n+1}}{S_{n}}\right|= \left|x+1\right|[/tex]

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 [tex]\sum^{+\infty}_{n=1} \frac{1}{\sqrt{n}}[/tex] (with p=1/2).

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

S

_{n}= (1)^{n+1}1/√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.