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Find radius of convergence for

$$\sum_{n=1}^{\infty} \frac{(x-1)^{n-3} + (x-1)^{n-1}}{4^n + 2^{2n-1}}$$

What I did:

$$\sum_{n=1}^{\infty} \frac{(x-1)^n ((x-1)^{-3} + (x-1)^{-1})}{2^{2n}(1+2^{-1})}$$

$$= \sum_{n=1}^{\infty} \frac{2}{3} ((x-1)^{-3} + (x-1)^{-1}) (\frac{x-1}{4})^n$$

$$L = |\lim_{x \to \infty} \frac{x-1}{4}| = \infty$$

$$R = L^{-1} = 0$$

But correct answer is 4

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I believe your limit expression is incorrect x does not go to infinity n does. Also as radius = reciprocal of what u calculated and point out u do get 4 for all value of x except when x = 1. – Comic Book Guy Mar 12 '12 at 3:38
up vote 1 down vote accepted

The way it's written, it's a weird series to ask for radius of convergence, as it has a singularity at $x=1$.

Forgetting about that, Alex Becker's answer is spot on: to find the radius of convergence you take the limit over $n$, not $x$, and only of the coefficients: here it would be $$ R=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim_{n\to\infty}\frac{\left(\frac14\right)^n}{\left(\frac14\right)^{n+1}}=4. $$

Or, if you want to go for understanding instead of just methods, again as Alex said this is a geometric series so it will converge when $|x-1|/4<1$, i.e. $$ |x-1|<4. $$

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Why do you take $\lim\limits_{x\to \infty}$? What you need to find is where $\sum_{n=1}^{\infty} \frac{2}{3} ((x-1)^{-3} + (x-1)^{-1}) (\frac{x-1}{4})^n$ converges, or since $\frac{2}{3} ((x-1)^{-3} + (x-1)^{-1})$ can be isolated from the series, equivalently find where $\sum\limits_{n=1}^\infty\left(\frac{x-1}{4}\right)^n$ converges. This is a geometric series, so you should be able to recognize when it converges.

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