What is the interval of convergence for this power series?

[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.

 

[Hurkyl]

Did you remember to take the limit?

 

[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} &lt; 1

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

 

[oldunion]

ah yes, that's it! i forgot the limit. thank you