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Example:

$\sum_{n=1}^{\infty} \frac{ 2x^{n+1}}{3^{n+1}*n^{3} }$

To find the ROC I used the ratio test:

I worked it down to this point where I wasn't sure entirely what to do:

$\lim_{n\to\infty} \bigg| \frac{2x}{3(n+1)} * \frac{n^{3}}{1} \bigg|$

$(\frac{1}{3})\lim_{n\to\infty} \bigg| \frac{2x}{(n+1)^{3}} * \frac{n^{3}}{1} \bigg|$

Then not knowing what to do I did this:

$(\frac{1}{3})\big|2x\big|\lim_{n\to\infty} \bigg|\bigg(\frac{n}{n+1} \bigg)^{3}\bigg|$

Then solving was unsure what I was to do with the |2x| should the 2 be divided out, if not, why?

$|2x|(\frac{1}{3}) < 1$

$|2x| < 3$

$|x| < \frac{3}{2}$ = radius of convergence

However, wolfram states this as ROC: 3

What am I doing wrong here?

Please help

Thank you

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  • $\begingroup$ Any factor that does not depend on n can be factored out before taking the limit. $\endgroup$ – user247327 Apr 27 '17 at 15:02
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The general term of the series is

$$a_n=\frac{2x^{n+1}}{3^{n+1}\,n^3}$$

The ratio test requires that we find that limit $\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$.


In the OP, a factor of $2$ common to $a_n$ and $a_{n+1}$ should have cancelled upon taking the ratio $\frac{a_{n+1}}{a_n}$


Proceeding to apply the test we have

$$\begin{align} \lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to \infty}\left|\frac{\frac{2x^{n+2}}{3^{n+2}\,(n+1)^3}}{\frac{2x^{n+1}}{3^{n+1}\,n^3}}\right|\\\\ &=\lim_{n\to \infty}\left|\frac{xn^3}{3(n+1)^3}\right|\\\\ &=\frac13|x| \end{align}$$

The series converges when the limit is less than $1$, diverges when the limit exceeds $1$, and is otherwise inconclusive when the limit is $1$.

Therefore, we see that the series converges when $|x|<3$.

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The ratio test gives $\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{2x^{n+2}3^{n+1}\cdot n^3}{2x^{n+1}3^{n+2}(n+1)^3}|=\frac {|x|}3$. For our radius of convergence we need $\frac {|x|}3=1$, i.e. $|x|=3$

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