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I have the following power series and I would like to figure out the radius of convergence:

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

I appreciate any help&explanation.

Jacky

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  • $\begingroup$ Are you aware of any test you could use to do so? $\endgroup$ – user88595 Feb 6 '14 at 14:45
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    $\begingroup$ Your power series is also known as $\cos(x)-1$. It converges on the entire complex plane. $\endgroup$ – J.R. Feb 6 '14 at 14:52
  • $\begingroup$ What is exactly that plane numerically? $\endgroup$ – Jacky Feb 6 '14 at 15:08
  • $\begingroup$ @Jacky: Don't worry about it. The upshot is that it converges everywhere (has infinite radius of convergence). See my answer for a hint at how to show this. $\endgroup$ – Cameron Buie Feb 6 '14 at 15:10
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$$\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|={}\lim\limits_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(x)^{2n+2}}{(2n+2)!}}{(-1)^n\frac{(x)^{2n}}{(2n)!}}\right|=\lim\limits_{n\to\infty}\frac{|x|^{2}}{(2n+1)(2n+2)}=0 \quad for \ all \ x\in \mathbb{R}.$$ So it is convergent for all $x\in \mathbb{R}$.

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Hint: Make the substitution $t=x^2,$ so that the series becomes $$\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}t^n.$$ Now try using the ratio test to determine for which $t$ this converges. That will tell you for which $x$ this converges.

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