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Find the radius of convergence of each of the following power series:

a.) $\sum_{j=1}^{\infty}\frac{2^j}{3^j+4^j}z^j$

b.) $\sum_{j=0}^{\infty}2^jz^{j^2}$

c.) $\sum_{j=0}^{\infty}[2+(-1)^j]^jz^j$

For a: I can find the radius of convergence of the power series by the theorem that states $R=\frac{1}{\alpha}$ thus $\alpha= \lim_{n\rightarrow\infty}sup[\frac{2^j}{3^j+4^j}]^{1/j}=2$ thus $R=1/2$

For b: Similar to a and found that $\lim_{n\rightarrow\infty}sup=1$ thus $R=1/1=1$

For c: Similar and found that $\lim_{n\rightarrow\infty}sup=3$ thus $R=1/3$

Is this correct?

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up vote 1 down vote accepted

You confuse me a lot at a). The radius of convergence is $$R=\frac{1}{\limsup_{n\rightarrow \infty} \sqrt[n]{|a_n|}}$$ As your series is approximate something like $\frac{2^k}{4^k}$ the radius of convergence will be $2$.

For the proof use that $$\frac{1}{2^k}=\frac{2^k}{4^k} \geq \frac{2^k}{4^k +3^k } \geq \frac{2^k}{2 \cdot 4^k}=\frac{1}{2} \cdot \frac{1}{2^k}$$

For b) the limsup should be more something like $\sqrt{2}$

For c) i got the same.

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I see my mistake now in part a. You used the comparison test correct? – Q.matin Mar 5 '13 at 21:27
yes for the $n-$th root I used the comparison test – Dominic Michaelis Mar 5 '13 at 21:29
Thanks a lot, Dominic! – Q.matin Mar 5 '13 at 21:45
Dominic, I am a bit confused. Why did you compare the limit to $\frac{2^k}{4^k}$, why couldn't you instead compare it to $\frac{2^k}{3^k}?$ – Q.matin Mar 11 '13 at 5:44

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