Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=0}^{\infty}(-2)^nx^{2n}$.
Explanation of Method
Writing the series as $\sum(-2x^2)^n$ reveals it as a geometric series with ratio $r=-2x^2$. It converges exactly when $|r|<1$, i.e. $|x|<1/\sqrt2$, and diverges at both endpoints where $|r|=1$ makes it the oscillating series $\sum(-1)^n$.
- Recognize the series as geometric in $x^2$, with ratio $-2x^2$.
- Solve $|-2x^2|<1$ for the radius.
- Test both endpoints, where the ratio has magnitude exactly $1$.
Step 1 · Recognize the geometric series
$$ \sum_{n=0}^{\infty}(-2)^nx^{2n} = \sum_{n=0}^{\infty}(-2x^2)^n, $$ a geometric series with ratio $r=-2x^2$, converging iff $|r|<1$.
Step 2 · Radius of convergence
$$ |-2x^2|<1 \iff x^2<\frac12 \iff |x|<\frac1{\sqrt2} \implies R=\frac1{\sqrt2}=\frac{\sqrt2}2. $$
Step 3 · Endpoints $x=\pm1/\sqrt2$
At either endpoint, $x^2=1/2$, so $r=-2\cdot\frac12=-1$, and the series becomes $$ \sum_{n=0}^{\infty}(-1)^n = 1-1+1-1+\cdots, $$ whose partial sums oscillate between $0$ and $1$ without settling — it diverges. This happens identically at both $x=1/\sqrt2$ and $x=-1/\sqrt2$, since only $x^2$ appears.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1/\sqrt2$ |
| At $x=\pm1/\sqrt2$ | Diverges (oscillating $\sum(-1)^n$) |
| Interval of convergence | $(-1/\sqrt2,\,1/\sqrt2)$ |
| Closed form | $1/(1+2x^2)$ |
Additional Notes
- Since only even powers of $x$ appear, the function is automatically even ($f(-x)=f(x)$), which is why both endpoints behave identically — a structural feature not present in Problems 223–228, all of which involve every power of $x$ and typically show asymmetric endpoint behavior.
- This is the negative-ratio geometric-series counterpart of Problem 209's $\sum(x/(x+1))^{\alpha n}$: both problems reduce to a pure geometric series in a transformed variable, and both diverge at any boundary point where the transformed ratio has magnitude exactly $1$.
- The closed form $1/(1+2x^2)$ is a smooth, bounded, even function on all of $\mathbb R$ — the series representation is valid only on the much smaller interval $(-1/\sqrt2,1/\sqrt2)$, illustrating that a power series' domain can be far smaller than the domain of the analytic function it represents.