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The following question was used as an example in my calc textbook:

Find the radius of convergence and interval of convergence of the series:

$\sum_{n=0}^\infty \frac{(-3)^nx^n}{\sqrt{n+1}}$

In the example, they use the ratio test to determine that the series converges when |x|<$\frac{1}{3}$. My question is, doesn't the ratio test for absolute convergence? I understand that absolute convergence implies convergence, but since absolute convergence is stronger than convergence, wouldn't there be values of x in an alternating series where the series is convergent but not absolutely convergent, and therefore this answer only gives part of the solution? Or is that taken into account by testing the end-points of the interval?

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  • $\begingroup$ The only values you need to check additionally would be $x=\pm \frac13$, the rest are covered by ratio / root tests. $\endgroup$ – Macavity May 3 at 15:26
  • $\begingroup$ A power series can only be conditionally convergent at the endpoints of its interval of convergence. See here. $\endgroup$ – kccu May 3 at 15:54
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You have correctly observed that the ratio test shows you absolute convergence and hence convergence. However it's a little inaccurate to call a power series alternating, because it can depend on the value of $x$. In your example $$ \sum \frac{(-3)^n x^n}{\sqrt{n+1}}$$

the power series is alternating if $x\geq0$, however if $x<0$ then the negatives from the $x$ and the negatives from the $-3$ cancel. As you can see, it only makes sense to take about a power series being alternating for a particular value of x.

For your particular series $x=1/3$ gives an alternating series that is conditionally convergent, $0 \leq x \leq 1/3$ gives alternating series that are absolutely convergent, and $x>1/3$ gives series which are alternating but divergent.

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