# INT of convergence problem

#### orendacl

(-1)^n(x+2)^n
______________
sqrt(n)*3^n

i use the ratio test abs(bn+1/bn), then plug and chug and I get:

sqrt(n)|x+2|
___________
3*sqrt(n+1)

=
|x+2|
________
3

the problem is, i try maple and it gets absolutely no where near that...rather, it gets infinity.???
I need to be able to find the int of convergence. thanks

#### Prove It

MHF Helper
(-1)^n(x+2)^n
______________
sqrt(n)*3^n

i use the ratio test abs(bn+1/bn), then plug and chug and I get:

sqrt(n)|x+2|
___________
3*sqrt(n+1)

=
|x+2|
________
3

the problem is, i try maple and it gets absolutely no where near that...rather, it gets infinity.???
I need to be able to find the int of convergence. thanks
I take it that this is actually

$$\displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^n(x + 2)^n}{3^n\sqrt{n}}$$.

The series converges where

$$\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1$$

$$\displaystyle \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}(x + 2)^{n + 1}}{3^{n + 1}\sqrt{n + 1}}}{\frac{(-1)^n(x + 2)^n}{3^n\sqrt{n}}}\right| < 1$$

$$\displaystyle \lim_{n \to \infty}\left|\frac{\frac{(x + 2)^{n + 1}}{3^{n + 1}\sqrt{n + 1}}}{\frac{(x + 2)^n}{3^n\sqrt{n}}}\right| < 1$$

$$\displaystyle \lim_{n \to \infty}\left|\frac{3^n(x + 2)^{n + 1}\sqrt{n}}{3^{n + 1}(x + 2)^n\sqrt{n + 1}}\right| < 1$$

$$\displaystyle \lim_{n \to \infty}\left|\frac{(x + 2)\sqrt{n}}{3\sqrt{n + 1}}\right| < 1$$

$$\displaystyle \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n + 1}} < 1$$

$$\displaystyle \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\sqrt{\frac{n}{n + 1}} < 1$$

$$\displaystyle \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\sqrt{1 - \frac{1}{n + 1}} < 1$$

$$\displaystyle \left|\frac{x + 2}{3}\right|(1) < 1$$

$$\displaystyle \left|\frac{x + 2}{3}\right| < 1$$

$$\displaystyle -1 < \frac{x + 2}{3} < 1$$

$$\displaystyle -3 < x + 2 < 3$$

$$\displaystyle -1 < x < 5$$.

So the interval of convergence is

$$\displaystyle -1 < x < 5$$.

You will need to check the endpoints though, as the ratio test tells us nothing when the limit = 1...

#### alexmahone

The interval of convergence is (-5, 1).

$$\displaystyle -3 < x + 2 < 3$$

$$\displaystyle -1 < x < 5$$
Say what?

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#### orendacl

I agree with member 'prove it'.
I do get ]x+2]/3 which is then (-1, 5) for int of convergence

#### 11rdc11

The interval of convergence is (-5, 1).

I get what you get as the interval of convergence.

$$\displaystyle -3 < x+2 <3$$

Now subtract 2 form both sides and you get

$$\displaystyle -5 < x < -1$$

Here is a different approach to this problem i was taught in my DE book.

$$\displaystyle \displaystyle{\lim_{n \to \infty} \frac{\frac{1}{3^{n+1}\sqrt{n+1}}}{\frac{1}{3^n\sqrt{n}}}}$$

$$\displaystyle \lim_{n \to \infty}\frac{3^n\sqrt{n}}{3^{n+1}\sqrt{n+1}} = \frac{1}{3}$$

Now take L= $$\displaystyle \frac{1}{3}$$

and p = $$\displaystyle \frac{1}{L}$$

and is p is equal to radius of convergence which is 3

so now

$$\displaystyle -3 < x +2 < 3$$

$$\displaystyle -5 < x < 1$$

when x = -5, the series is

$$\displaystyle \frac{1}{\sqrt{n}}$$ which diverges by the p test

when x = 1, the series is

$$\displaystyle \frac{(-1)^n}{\sqrt{n}}$$ which converges through the alternating series test

so

$$\displaystyle -5 < x \leq 1$$

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