Interval of convergence

Jul 2008
46
0
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

Feb 2010
240
64
China
Post your work to see whats wrong with it.
PS: the Root Test is better for this one.

and remember that: \(\displaystyle \lim_{n\to\infty} n^{\frac{1}{n}} = 1\).
 
Dec 2009
1,506
434
Russia
...

Solution with test root:

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


So, the interval of convergence is:

\(\displaystyle \frac{-1}{6}<x+2<\frac{1}{6}\)
\(\displaystyle \frac{-13}{6}<x<\frac{-11}{6}\)

NOW CHECK CONVERGENCE FOR EDGE POINTS:
 
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