Interval of convergence of the power series Given $$\sum_{n=1}^\infty  \frac{(-1)^n}{4^nn^p}x^{(2n)}$$
find the interval of convergence.
Using the ratio test, I ended up with the following assuming that $$ x^{(2n)} $$ is part of the numerator.
$$\lim_{n\to \infty} \vert x^2 \vert\frac{1}{4}\left(\frac{n}{n+1}\right)^p$$
This looks like a Geometric series. I am not sure if this is correct. I am also struggling to deal with the $$\frac{1}{4}\left(\frac{n}{n+1}\right)^p$$.
I have tried this
because $$\left(\frac{n}{n+1}\right)^p $$ looks like an Indeterminate Form.
the reciprocal is: $$\frac{1}{\left(\frac{n+1}{n}\right)^p}$$
$$e^{\lim_{n\to \infty} ln{\left(\frac{n+1}{n}\right)^p}}$$
$$e^{\lim_{n\to \infty} p ln{\left(\frac{n+1}{n}\right)}}$$
$$e^{\lim_{n\to \infty} \frac{ln{\left(\frac{n+1}{n}\right)}}{\left(\frac{1}{p}\right)}}$$
L'Hopital
$$\frac{1}{\left(\frac{n+1}{n}\right)}$$
$$e^{\lim_{n\to \infty} \frac{\}\frac{1}{\left(\frac{n+1}{n}\right)}{-\frac{1}{p^2}}}{-\frac{1}{p^2}}}  $$ 
the P's cancel each other, therefore
$$e^{\lim_{n\to \infty} \frac{1}{\left(\frac{n+1}{n}\right)} = {-\frac{1}{e}} } $$
Is this correct?
 A: The root test actually gives:
$$\lim_{n \to \infty}\left(\frac{(-1)^n}{4^nn^ p}x^{2n}\right)^{1/n} = \lim_{n \to \infty} x^2\cdot\frac{1}{4}\cdot\left(\frac{1}{n^p}\right)^{1/n} = \frac{x^2}{4}\lim_{n \to \infty} \left(\frac{1}{n^{1/n}}\right)^p$$
Is this clear? Can you solve it from here? (remember that $x \mapsto \left(\frac{1}{x}\right)^p$ is continuous for $x \neq 0$)
EDIT: For it to be a properly power series in $x$, then $p$ needs to be a fixed parameter: for each $p$, we consider the power series in $x$ given by:
$$\sum_{n = 1}^\infty \frac{(-1)^n}{4^nn^p}x^{2n}$$
Now, for each fixed value of $p$, we want to determine the radius of convergence of the series above. It may or it may not depend on the value of $p$ chosen (hint: it doesn't).
When calculating the limit, you considered wheter
$$\left(\frac{n}{n+1}\right)^p$$
was an indeterminate form. It really evaluates to $\infty/\infty$, so you could apply the L'Hopital rule. But remember that we are calculating the limit with respect to $n$, leaving $p$ fixed. When you applied the rule, you differentiated part of it with respect to $n$, and part of it with respect to $p$. L'Hopital's rule states that:
$$\lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{\frac{df}{dn}(n)}{\frac{dg}{dn}(n)}$$
whenever $f$ and $g$ are differentiable functions of $n$ and $f(n)/g(n)$ is an indeterminate form. Our functions are: $f(n) = n^p$ and $g(n) = (n+1)^p$. So L'Hopital gives:
$$\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^p = \lim_{n \to \infty} \frac{pn^{p-1}}{p(n+1)^{p-1}}$$
It is possible to solve this limit trough $p$ repeated uses of the rule, but it is simpler to calculate the limit above if you notice that you can rewrite the denominator:
$$n + 1 = n\left(1 + \frac{1}{n}\right)$$
And that $x \to \left(\frac{1}{x}\right)^p$ is continuous for every fixed $p$ and $x \neq 0$. For a continuous function $f$ and any function $g$, we have: $\lim_{x \to ?} f(g(x)) = f(\lim_{x \to ?} g(x))$ (taking care that everything is defined). When making calculus with several incognitos, try to keep clear what is being treated as a variable and what is being treated as a fixed parameter.
