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Analyze the convergence behavior of the following series: $$\sum_{k=0}^{\infty}\frac{x^{2k}}{2^{2k}}-\frac{x^{2k+1}}{3^{2k+1}}.$$

I came across this problem as I was preparing for an exam. It is supposed to be an easy one but I am not sure which test to apply.

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Hint: determine for which values of $x$ the general term of the series converges to zero. Or use the criterion based on the limsup of the $n$th root. –  Did May 27 '12 at 21:21
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2 Answers

up vote 3 down vote accepted

I assume that $$\sum_{k=0}^{\infty}\left(\frac{x^{2k}}{2^{2k}}-\frac{x^{2k+1}}{3^{2k+1}}\right)$$ is intended. You can avoid worrying about uniform convergence by using the ratio test. Calling the $k$-th term $a_k$, and letting $u=x/2$, we have

$$a_k=\left(\frac{x}2\right)^{2k}-\left(\frac{x}3\right)^{2k+1}=u^{2k}-\left(\frac23\right)^{2k+1}u^{2k+1}=u^{2k}\left(1-\left(\frac23\right)^{2k+1}u\right)\;,$$

so

$$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}u^2\left|\frac{1-(2/3)^{2k+3}u}{1-(2/3)^{2k+1}u}\right|=u^2\;.$$

Clearly you need $u^2<1$, or $|x|<2$, and it’s not hard to check that you don’t get convergence at either endpoint.

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Break it in two using the - sign. Then examine the pieces using the root or ratio test. More simply, realize that both series are geometric, which converges when the common ration is less than 1 in absolute value.

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but to break it don't we need to know that the series converges uniformly? –  Galois May 27 '12 at 21:28
    
It has to converge absolutely. Power series converge absolutely in the interior of their intervals of convergence. Find the intervals of convergence; one fits inside the other. Then you can check at the smaller interval's endpoints separately. This will tell all. –  ncmathsadist May 27 '12 at 21:30
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