Interval of convergence

**Casas4** · June 24th 2010, 10:48 PM

Find the interval of convergence for the following power series:

I used the ratio test first to get my radius of convergence which I got to be 6 but I think this is where I may have messed up. From there I added 6 to and subtracted 6 from 2, which would make my interval (-4,8) but its not right so can anyone help?
Thanks
AC

**Ted** · June 24th 2010, 10:57 PM

Post your work to see whats wrong with it.
PS: the Root Test is better for this one.

and remember that: $\lim_{n\to\infty} n^{\frac{1}{n}} = 1$ .

**Also sprach Zarathustra** · June 25th 2010, 04:00 AM

Solution with test root:

$lim_{n\to\infty} \sqrt[n]{\frac{|(x+2)^n|{6^n}}{7n^4}} = lim_{n\to\infty} \frac{6|(x+2)|}{ \sqrt[n]{7n^4}}=\frac{6|x+2|}{lim_{n\to\infty}\sqrt[n]{7n^4}}=\frac{6|(x+2)|}{1}<1$

So, the interval of convergence is:

$\frac{-1}{6}<x+2<\frac{1}{6}$
$\frac{-13}{6}<x<\frac{-11}{6}$

NOW CHECK CONVERGENCE FOR EDGE POINTS:

**Ted** · June 25th 2010, 04:08 AM

^^^
Your solution is wrong.

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