Series question:

Question

Series question:

anonymous · Answer

Find the radius of convergence:

$$\huge \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$$
So if the sequence of the series converges, then the series also converges
$$\huge \lim_{n \rightarrow \infty}\frac{ x^{2n} }{ n! }$$
Now here's what the solution says:
$$\huge \lim_{n \rightarrow \infty} \sqrt[n]{\frac{ x^{2n} }{ n! }}=x^2 \lim_{n \rightarrow \infty}\frac{ 1 }{ \sqrt[n]{n!} }$$
I don't understand why we're allowed to take the nth root here..

amistre64 · Answer

nth roots, for n>1, are monotonic increasing might have something to do with the 'why'

amistre64 · Answer

i usually just work a ratio thing

anonymous · Answer

Are you referring to the ratio test?

amistre64 · Answer

i might be, i never did get the terminology down pat

amistre64 · Answer

$$\lim_{n\to \infty}{\frac{a_{n+1}}{a_n}}$$

anonymous · Answer

Yeah, okay. That makes sense. But I still don't really understand why they took the nth root instead. I understand the purpose to remove the nth power from x^2, but I don't understand *why* we can do that.

amistre64 · Answer

http://tutorial.math.lamar.edu/Classes/CalcII/RootTest.aspx prolly explains it better than i can

anonymous · Answer

Oh geez, duh. I didn't realize they were using the root test -- there was no justification in their steps. Perfect! Thanks

amistre64 · Answer

youre welcome