Question

Find the interval of convergence of the power series (see attached problem)

Answer 1

\[\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!}\]
Apply the ratio test:
\[\begin{align*}\lim_{n\to\infty}\left|\frac{(-1)^{n+1}x^{2(n+1)}}{(n+1)!}\cdot\frac{n!}{(-1)^nx^{2n}}\right|&=\lim_{n\to\infty}\left|\frac{(-1)^{n}(-1)x^{2n}x^2}{(n+1)n!}\cdot\frac{n!}{(-1)^nx^{2n}}\right|\\
&=\lim_{n\to\infty}\frac{x^2}{(n+1)}\\
&=0
\end{align*}\]
What does this tell you?

Answer 2

series converges because 0<1

Answer 3

Right, so if it converges *regardless* of the values of \(x\), what's your radius of convergence?

Answer 4

the interval of convergence: \[-\infty < x < \infty \] ?

Answer 5

Yes, that's the one!

Answer 6

awesome! thanks for your help