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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Sign up to join this communityI 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
$$\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|={}\lim\limits_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(x)^{2n+2}}{(2n+2)!}}{(-1)^n\frac{(x)^{2n}}{(2n)!}}\right|=\lim\limits_{n\to\infty}\frac{|x|^{2}}{(2n+1)(2n+2)}=0 \quad for \ all \ x\in \mathbb{R}.$$ So it is convergent for all $x\in \mathbb{R}$.
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.