# Thread: INT of convergence problem

1. ## INT of convergence problem

(-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

2. Originally Posted by 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
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...

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

Originally Posted by Prove It
$\displaystyle -3 < x + 2 < 3$

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

4. ## reply regarding int of conv.

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

5. Originally Posted by alexmahone
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.

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

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

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

$\displaystyle -5 < x \leq 1$