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Given

$\sum_{k=0}^\infty$ $(-1)^k$ $(2x)^{2k}$,

I have to evaluate the interval of convergence.

Approach

I have to evaluate the convergence radius. Using Cauchy-Hadamard, I receive:

$\lim_{k\to \infty}$ $\sqrt[k]{1 \over |-1|^k}$ $=$ $1 \over |-1|$,

and therefore, we have to solve

$|2x|^2 \over |-1|$ $< 1$ $\Rightarrow$ $|2x|^2$ $< 1$ $\Rightarrow |2x| < 1$ $\Rightarrow -1 < 2x < 1$ $\Rightarrow$ $-1 \over 2$ < $x$ $<$ $1 \over 2$.

It follows that the interval of convergence is at least as large as

$( $$ 1\over 2$, $1 \over 2$).

Now, the behaviour of the series at the points on the edge could be evaluated further.

Community
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Julian
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1 Answers1

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Using the ratio test, we obtain $$\lim_{k\to\infty} \frac{(-1)^{k+1}(2x)^{2k+2}}{(-1)^{k}(2x)^{2k}} = -(2x)^2$$ And so we want $$-1 < 4x^2 < 1$$ $$-1/4 < x^2 < 1/4$$ $$-1/2 < x < 1/2$$

To test the endpoints, simply substitute $x = \frac{1}{2}$ and $x = \frac{-1}{2}$ into the series. We get, respectively, $$\sum_{k=0}^{\infty}(-1)^k\cdot (1)^{2k}$$ $$\sum_{k=0}^{\infty}(-1)^k\cdot (-1)^{2k}$$ Neither of which converge. So the interval of convergence is merely $(-1/2, 1/2)$.

florence
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  • Thanks - we have the same solution then. :-) Would you mind taking a look at my solution? You solved this by the radio test, I solved it with the help of Cauchy-Hadamard. But did I do it right or are there some flaws? Sure, the result is the same, but there might be flaws in my representation. – Julian Jul 04 '16 at 08:34
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    @Julian It seems good to me. – florence Jul 04 '16 at 08:37