(-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...