<aside>

Find a power series representing the function $f(x) = \frac{2}{x+3}$ near 0, and then find its radius/interval of convergence.

</aside>

To do this, we can use the geometric series formula $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$.

We can rewrite $f(x)$ to get it to look like the right-hand side.

$$ f(x) = 2\left(\frac{1}{3-(-x)}\right) = \frac{2}{3}\left(\frac{1}{1-(-x/3)}\right) $$

So $f(x) = \displaystyle\sum_{n=0}^\infty\frac{2}{3}\left(-\frac{x}{3}\right)^n = \sum_{n=0}^\infty\frac{2(-1)^n}{3^{n+1}}x^n$.

This series converges when $|\frac{x}{3}| < 1$ (using the geometric series test), which happens when $|x| < 3$.

The radius of convergence is $R = 3$, and the interval of convergence is $(-3, 3)$.

<aside>

Find a power series representing the function $f(x) = \frac{2}{x+3}$ near 0, and then find its radius/interval of convergence.

</aside>

$$ f(x) = \frac{1}{1-(-x^2)} = \sum_{n=0}^\infty(-x^2)^n = \sum_{n=0}^\infty(-1)^n(x^{2n}) $$

By the alternating series test, the series converges when $|x| < 1$, so the radius of convergence is $R=1$ and the interval of convergence is $(-1, 1)$.