Radius of convergence of $\sum_{n=1}^{\infty} (3^n +(-5)^n) x^{7n}$ What is the radius of convergence of the power series $\sum_{n=1}^{\infty} (3^n +(-5)^n) x^{7n}$?
I tried to find the limit of $\dfrac{a_{n+1}}{a_n}$ and I found it is $5$. Is that true, and does $x^{7n}$ play any role?
 A: Change variable $y=x^7$, it would be easy to see the sum converges on 
$$
-\frac{1}{5}< y <\frac{1}{5}\iff-\frac{1}{\sqrt[7]{5}}< x <\frac{1}{\sqrt[7]{5}}
$$
So the convergence radius is $\frac{1}{\sqrt[7]{5}}$.
A: It's perhaps a little better to take $a_n=(3^n+(-5)^n)x^{7n}$ (or do as Melting Snow suggests and do a variable change).  Then
\begin{align*}
\lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_n}\right| &= 
|x|^7\lim_{n\rightarrow\infty}  \left|\frac{3^{n+1}+(-5)^{n+1}}{3^n+(-5)^n}\right|
\end{align*}
Dividing top and bottom of the fraction by $(-5)^n$, we get
$$\lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_n}\right| = 5|x|^7,$$
and this is less than $1$ if and only if $\displaystyle{|x|<\frac{1}{\sqrt[7]{5}}}=R$.
As a bonus, for $|x|=R$ we can easily see that $|a_n|\rightarrow 1$, and so the series diverges everywhere on the boundary.

More generally, if you have a power series $\sum b_n x^{kn}$ and $\displaystyle{\limsup_{n\rightarrow\infty} \left| \frac{b_{n+1}}{b_n}\right| = \frac{1}{S}}$, then the domain of absolute convergence is $|x| < \sqrt[k]{S}$.  Similarly for the root test or any other test of absolute convergence based on the coefficients.  In your case, you showed that $S=1/5$, and then taking the 7-th root you get the answer above.
