Question

I'm having some difficulty applying the root test to discover the ratio of convergence for a power series. (Posted below in a minute).

Answer 1

\[\sum_{n = 0}^{\infty }\frac{nx^{n}}{n+2}\]

Answer 2

\[\left| \frac{ \sqrt[n]{nx^{n}} }{ \sqrt[n]{n+2} } \right|\]

Answer 3

Is this correct so far or have I misused a property of exponents or something? @agent0smith

Answer 4

(Sorry, LaTeX not loading properly atm, trying to continue it)

Answer 5

\[\left| \frac{ nx^{n/n} }{ \sqrt[n]{n+2} } \right| = \left| \frac {nx}{\sqrt[n]{n+2}} \right|\]

Answer 6

Is this all okay so far?

Answer 7

You're missing the \(n\)-th root on the \(n\) in the numerator.

Answer 8

Oh, yep, thanks. \[\left| \frac{ \sqrt[n]{n}x }{ \sqrt[n]{n+2} } \right| \]\[\lim_{n \rightarrow \infty}\left| \frac{ \sqrt[n]{n}x }{ \sqrt[n]{n+2} } \right| = \left| \frac{ x }{ 3 } \right|\]

Answer 9

Not so sure about that last denominator.

Answer 10

I think the nth root of (n+2) (as n approaches infinity) is still 1.

Answer 11

Ty, I'm gonna check using W|A or something, first time I've run into this

Answer 12

Yep, it does.

Answer 13

Okay, so it'd just be abs|x|, right?

Answer 14

I get the feeling ratio test would make for easier algebra. Are you required to use the root test?

Answer 15

Nah, just seeing if I can do it this way, no requirements on how we discover the radius of convergence/domain of x where x converges.

Answer 16

Okay , sweet, this was right, thanks for catching the nth root in the numerator, I didn't see that and was really confused as to how to do this.

Answer 17

yw