Find the radius of convergence of the following power series;
$$\sum_{j = 0}^\infty\frac{2^j}{3^j+4^j}(z^j)^2$$
Now I'm trying to do a root test, but I don't know how to apply it here since it becomes a problem in the denominator, I don't know how to take the root of j out of two terms.
First consider the subsituted eqatuion, with $z^2=x$. I also replaced $j$ by $n$:
$$\sum_{n = 0}^\infty\frac{2^n}{3^n+4^n}(x)^n$$
The ratio test:
$$R_x=\lim_{n\to \infty}|a_n/a_{n+1}|=0.5\lim_{n\to \infty}\frac{3^{n+1}+4^{n+1}}{3^{n}+4^n}$$.
Now divide the expand the fraction with $4^{-n-1}$:
$$R_x=0.5\lim_{n\to \infty}\frac{(0.75)^{n+1}+1}{1/3(0.75)^{n+1}+0.25^\cdot }=2$$.
In the previous limit all terms $(0.75)^n \to 0$ as $n\to \infty$. We are left with $1/0.25=4$. So the radius of convergence for x is $R_x=2$. Take the squareroot to get the radius of convergence for $z$ to be $R_z=\sqrt{2}$.
If we express your series as $\sum\limits_{j = 0}^\infty a_j z^j$, then $$a_j = \begin{cases}\dfrac{2^{j/2}}{3^{j/2} + 4^{j/2}}& \text{if $j$ is even}\\0&\text{otherwise}\end{cases}.$$
Since $\lim\limits_{j\to \infty} \lvert a_{2j-1}\rvert^{1/(2j-1)} = 0$ and
$$\lim_{j\to \infty} \lvert a_{2j}\rvert^{1/{2j}} = \frac{2^{1/2}}{\lim\limits_{j\to \infty} (3^{j} + 4^{j})^{1/{2j}}} = \frac{2^{1/2}}{\max\{3,4\}^{1/2}} = \frac{1}{\sqrt{2}},$$
then $\limsup\limits_{j\to \infty}\, \lvert a_j\rvert^{1/j} = \dfrac{1}{\sqrt{2}}$. Hence, the radius of convergence is $\sqrt{2}$.
$$\frac{2^n(z^2)^n}{3^n+4^n}=\frac{2^n(z^2)^n}{4^n\left(1+(\frac{3}{4})^n\right)}=\frac{z^{2n}}{2^n\left(1+(\frac{3}{4})^n\right)}$$ now take the n-root $$L=\lim_{n\rightarrow \infty }\left(\frac{z^{2n}}{2^n\left(1+(\frac{3}{4})^n\right)}\right)^{\frac{1}{n}}=\lim_{n\rightarrow \infty }\frac{z^2}{2(1+(\frac{3}{4})^n)^{\frac{1}{n}}}=|\frac{z^2}{2(1)}|<1$$ so the
