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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.

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  • $\begingroup$ Note that $4^j$ dominates $3^j$. Also a caveat: the power term is $z^{2j}$ instead of $z^j$, so be careful to apply any ratio test (Cauchy test isn't affected, though.) $\endgroup$ – Vim Sep 29 '16 at 15:24
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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 $R_z=\sqrt{2}$.

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  • $\begingroup$ I understand your ratio test, but I can't say I understand how the last part was equal to 2. It's 0.5 times 4, but how does that last limit equal 4? $\endgroup$ – armara Sep 29 '16 at 15:48
  • $\begingroup$ As far as i remember $R=\lim_{n\to \infty}\frac{1}{|a_{n+1}/a_n|}=\lim_{n\to \infty}\frac{|a_n|}{|a_{n+1}}$ $\endgroup$ – MrYouMath Sep 29 '16 at 15:52
  • $\begingroup$ Yeah sorry, I deleted that comment. I'm just wondering about that last limit now, and how it equals 4. $\endgroup$ – armara Sep 29 '16 at 15:53
  • $\begingroup$ Sorry, I just got it. Clouded mind, thank you for the help! $\endgroup$ – armara Sep 29 '16 at 15:55
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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 $\sqrt{2}$.

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$$\frac{2^n(z^2)^n}{3^n+4^n}=\frac{2^n(z^2)^n}{4^n\left(1+(\frac{3}{4})^n\right)}=\frac{z^{2n}}{2^n\left(1+(\frac{3}{4})^n\right)}$$ 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

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