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


I appreciate any help&explanation.


  • $\begingroup$ Are you aware of any test you could use to do so? $\endgroup$
    – user88595
    Feb 6, 2014 at 14:45
  • 1
    $\begingroup$ Your power series is also known as $\cos(x)-1$. It converges on the entire complex plane. $\endgroup$
    – J.R.
    Feb 6, 2014 at 14:52
  • $\begingroup$ What is exactly that plane numerically? $\endgroup$
    – Jacky
    Feb 6, 2014 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. $\endgroup$ Feb 6, 2014 at 15:10

2 Answers 2


$$\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|={}\lim\limits_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(x)^{2n+2}}{(2n+2)!}}{(-1)^n\frac{(x)^{2n}}{(2n)!}}\right|=\lim\limits_{n\to\infty}\frac{|x|^{2}}{(2n+1)(2n+2)}=0 \quad for \ all \ x\in \mathbb{R}.$$ 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.


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