I'm having trouble with the algebraic manipulation required to find the interval of convergence of many of the Power series and Taylor series problems in my calculus class.

For example:

$\displaystyle \sum_{n=0}^{\infty}\left ( -1 \right )^{n}\frac{n}{4^{n}}x^{2n}$

Using the Ratio test, and ignoring $\displaystyle \left ( -1 \right )^{n}$ because we can, we have:

$\displaystyle \left | \frac{a_{n+1}}{a_{n}} \right|=\frac{(n+1)x^{(2n+1)}}{4^{n+1}}*\frac{4^{n }}{nx^{2n}}$

The $\displaystyle 4^{n}$ can be cancelled:

$\displaystyle \frac{(n+1)x^{(2n)}*x}{4}*\frac{1}{nx^{2n}}$

Distributing the $\displaystyle x^{2n}$:

$\displaystyle \frac{nx^{(2n)}+x^{(2n)}*x}{4}*\frac{1}{nx^{2n}}$

Cancelling the $\displaystyle nx^{2n}$ terms:

$\displaystyle =\frac{1+x^{(2n)}*x}{4}*\frac{1}{1}=\frac{1+x^{(2n +1)}}{4}$

The next step:

$\displaystyle \lim_{n \to \infty }\frac{1+x^{(2n+1)}}{4}$

I want to say that the limit is infinity but I think that would mean the interval of convergence is all real numbers; I don't think that is the answer. Perhaps I made algebraic/arithmetic mistake or perhaps I am not considering the limit correctly.