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$$\sum_1^\infty \frac{2^n}{3n}(x+3)^n$$

I do the ratio test to find the radius.

$$\frac{2^{n+1}}{3(n+1)}(x+3)^{n+1} *\frac{3n}{2^n (x+3)^n}$$

This should reduce down to $2|x+6|< 1$

This is wrong though, where did I go wrong?

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You should divide the terms instead of multiplying them. – Lord Soth Jul 27 '13 at 19:28
I did do that. I will edit. – Paul the Pirate Jul 27 '13 at 19:29
up vote 2 down vote accepted

Let $$u_n=\frac{2^n}{3n}(x+3)^n$$ then by the ratio test: $$\lim_{n\to\infty}\frac{|u_{n+1}|}{|u_n|}=2|x+3|<1\iff-\frac{7}{2} <x<-\frac{5}{2}$$ we can verify that the series is convergent at $x=-\frac{7}{2}$ (alternating series) so the interval of convergence is $[-\frac{7}{2},-\frac{5}{2})$.

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Where do you get 3? I get 6. – Paul the Pirate Jul 27 '13 at 19:40
Notice to find $u_{n+1}$ just replace $n$ by $n+1$ in $u_n$. – user63181 Jul 27 '13 at 19:42
The 3n is a linear term and doesn not "contribute" in ratiotest. When you consider the term (2x+6)^n by the rule of exponents, the then the by Sami states inequality becomes obvious, because by ratiotest, the exponent "drops". This reasoning is not "clean" to put on a quiz, nevertheless, it is true here. – imranfat Jul 27 '13 at 20:39

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