# Power series, driving me mad

1. May 4, 2005

### oldunion

summation from n=1 to infinity (x-2)^n/(n*3^n).

My teacher got -1 <= x<= 5 as the interval of convergence because he found that x-2/3<1

Using the ratio test i get (x-2)n/(n+1)3, consistantly. This is driving me wild. :rofl:

2. May 4, 2005

### Hurkyl

Staff Emeritus
Did you remember to take the limit?

3. May 4, 2005

### Jameson

$$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n3^n}$$

So using the ratio test we get,

$$\lim_{n\rightarrow\infty}\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}*\frac{n3^n}{(x-2)^n}$$

This reduces to

$$\lim_{n\rightarrow\infty}\frac{(x-2)n3^n}{(n+1)3^{n+1}}$$

Which further reduces to...

$$\lim_{n\rightarrow\infty}\frac{n(x-2)}{3(n+1)}$$

So, evaluating the limit we get:

$$\frac{{\mid}x-2{\mid}}{3} < 1$$

I think Hurkyl's advice was easier than my work.

Last edited: May 4, 2005
4. May 4, 2005

### oldunion

ah yes, thats it! i forgot the limit. thank you

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