Finding the limit of a quotient of polynomials containing a constant to the power of $n$ I've been given the infinite series:
$$ \sum_{n=0}^{\infty} (n+2^n)z^n $$
I need to find the convergence radius through:
$$R = \lim_{n \rightarrow \infty} \left| \frac{a_n}{a_{n+1}} \right|$$ 
My $a_n$ is $ a_n = n+2^n $ thus
$$R = \lim_{n \rightarrow \infty} \left| \frac{n+2^n}{(n+1)+2^{n+1}} \right|$$


*

*I'm having trouble using L'Hopital to find this limit, please show me how to do that.

*I tried ln like so:
$$\ln(R) = \lim_{n \rightarrow \infty} \left| \frac{\ln(n+2^n)}{\ln((n+1)+2^{n+1})} \right|$$
Am I allowed to insert the ln into the absolute value?
3 How do I use ln to solve this limit?


*I'm not sure, is this a Laurent series?


Many thanks.
 A: The limit can be solved by two applications of L’Hôpital’s. Here’s how:
$$\lim_{n\to\infty} \left|\frac{n+2^n}{n+1+2^{n+1}}\right| \\ =\lim_{n\to\infty} \left| \frac{1+2^n \ln 2}{1+2^{n+1} \ln 2} \right| \\ = \lim_{n\to \infty} \left| \frac{2^n (\ln 2)^2}{2\cdot 2^n (\ln 2)^2} \right| \\ =\frac 12 $$
And no, you’re not allowed to take the $\ln$ inside like that. Note that
$$\ln R = \ln\left(\lim_{n\to \infty} \left|\frac{n+2^n}{n+1+2^{n+1}} \right| \right) = \lim_{n\to \infty} \ln\left|\frac{n+2^n}{n+1+2^{n+1}}\right| \\ =\lim_{n\to \infty}\left[ \ln |n+2^n| - \ln|n+1+2^{n+1}|\right]$$
A: This is a Laurent series technically, centered at $0$, but one in which the coefficients for the $z^{-n}$ terms are zero. It is also a Taylor series and a Maclaurin series (since it's centered at $0$). 
Alternatively, it's handy to realize that the limit can be solved without l'Hopital's rule. As follows,
$$\lim_{n\rightarrow \infty} \left|\frac{n+2^n}{n+1+2^{n+1}}\right| = 
  \lim_{n\rightarrow \infty} \left|\frac{n/2^n + 1}{(n+1)/2^{n} + 2}\right| =
\lim_{n\rightarrow \infty} \left|\frac{0 + 1}{0 + 2}\right| = \frac{1}{2}.$$
