Originally Posted by

**xEnOn** To test for convergence of a series, I would usually take the limit of the series with root test's power 1/n.

But in some cases, I was told that I could use the power of the coefficient of a power series. For instance, in this series $\displaystyle {\textstyle \underset{p=2}{\sum}\frac{(4)^{2p}}{9^{p}p^{5/4}}(\frac{5}{4}-x)^{2p}}$, its coefficient has a power of 2p. I could just take the limit of the coefficient, $\displaystyle \underset{p\rightarrow\infty}{\lim}\frac{1}{\left[\frac{(4)^{2p}}{9^{p}p^{5/4}}\right]^{1/2p}}=\frac{3}{4}$. So 3/4 is the radius of convergence, which is correct. If I took the limit of the root test with power 1/n, the answer would be wrong.

But then in another series like this one: $\displaystyle \underset{n=2}{\sum}\frac{n^{n^{2}}e^{3n}}{(2+n)^{ n^{2}}2^{2n}}4^{2n}(x+\frac{5}{4})^{2n}$, when I take the limit of its coefficient with its power 2n, $\displaystyle \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\cdot e^{1/2}}$, the answer for the radius of convergence is wrong. The correct answer is 1/(4e). Apparently, I will only get the 1/(4e) as the answer if I took 1/n in my root test instead of 1/2n.

In this case, how would I know when I should take 1/n or 1/2n or even 1/(2n+1) in my root test for a series?

Thanks!