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Starting to learn convergence of a power series in complex analysis, and I was given two tests (and my lecture didn't differentiate between the usefulness of the two) which were the Ratio Test and the Root Test.

Information/Notes I have

For a complex power series
$$S = \sum_{n=0}^{\infty}a_n (z-z_0)^n$$

Denote the radius of convergence as $\rho$, then the root test tells us that
$$\rho = (\lim_{n\rightarrow \infty}\sup(|a_n|^\frac{1}{n})))^{-1}.$$

Then also (if it does exist), by the Ratio Test,
$$\rho=\lim_{n\rightarrow \infty} \left|\frac{a_n}{a_{n+1}}\right|$$
which is (to me) a much easier limit to find.
I read this: Radius of convergence and ratio test
and people say that if I could find $\rho$ by the ratio test, I could have done so by the Root Test as well. It seems much harder though, and I'm not sure if I understand it (I'm using W|A to calculate the supremum here).

Question in specific

$$S = \sum_{n=0}^\infty \frac{n(n+1)(n^2+2)}{2^n}(z+3)^n$$
The ratio test immediately shows that the radius of convergence $\rho$ is equal to $2$. However, I'm not sure how I would have used the Root Test here, and W|A doesn't seem to validate my answer by the Ratio Test... (http://www.wolframalpha.com/input/?i=supremum+of+(n(n%2B1)(n%5E2%2B2)%2F2%5En)%5E1%2Fn).
I was wondering if someone could clear up whether or not I'm supposed to be able to do it by the Root Test (even if it's computationally harder here) or did I just get some definition mixed up?

Twenty-six colours
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  • But $$\left(\frac{n(n+1)(n^2+2)}{2^n}\right)^{-1/n}\to2$$ so where is the problem? – Did Aug 01 '17 at 12:52
  • The choice of using ratio/root test is based on which would be computationally easier. You won't get different answers with both. Usually, root test is used when there is $n$ in the power. – Sahiba Arora Aug 01 '17 at 12:52
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    Oh, and of course, the root test always work while the ratio test works only in some restricted cases. – Did Aug 01 '17 at 12:54
  • @DId Why did you only look at the $|a_n|$ term? How come the $\lim \sup$ operation disappeared? – Twenty-six colours Aug 01 '17 at 12:56
  • @SahibaArora I was under the impression that some cases the Ratio Test wouldn't give as "accurate" (if any) answer than if we used the Root Test for some reason. e.g. in https://math.stackexchange.com/questions/331770/radius-of-convergence-and-ratio-test?noredirect=1&lq=1, the top reply, I did the ratio test and found the limit to be infinity, whilst they calculated it by the Root Test to be $3$. – Twenty-six colours Aug 01 '17 at 12:57
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    @Twenty-sixcolours Actually, I computed $$\lim|a_n|^{-1/n}$$ Of course, when this limit exists, it is also $$\frac1{\lim|a_n|^{1/n}}=\frac1{\limsup|a_n|^{1/n}}$$ but the main advantage of the root test is when the limit does not exist, only the limsup. – Did Aug 01 '17 at 13:00
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    @Twenty-sixcolours If the ratio test is applicable, it gives the same answer as the root test. But the ratio test is only applicable in special circumstances. – Daniel Fischer Aug 01 '17 at 13:01
  • @DanielFischer What kind of circumstances would they be? – Twenty-six colours Aug 01 '17 at 13:02
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    Only finitely many coefficients may be zero, and the limit of the quotients must exist. When the ratio test works, it's often simpler/easier to apply than the root test. But the root test always works, while the ratio test doesn't. – Daniel Fischer Aug 01 '17 at 13:03

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