Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{2^nx^n}{n^2+1}$.
Explanation of Method
The ratio test gives radius $1/2$. At either endpoint the series reduces to $\sum\pm1/(n^2+1)$ (or its alternating form), and since $1/(n^2+1)<1/n^2$, both endpoints converge absolutely by direct comparison — no delicate boundary analysis needed.
- Apply the ratio test to find $R$.
- Test $x=1/2$: compare $\sum1/(n^2+1)$ to $\sum1/n^2$.
- Test $x=-1/2$: the same comparison applies to the absolute values.
Step 1 · Radius of convergence
With $a_n=2^n/(n^2+1)$: $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = 2\lim_{n\to\infty}\frac{n^2+1}{(n+1)^2+1} = 2 \implies R=\frac12. $$
Step 2 · Endpoint $x=1/2$
$$ \sum_{n=1}^{\infty}\frac{2^n(1/2)^n}{n^2+1} = \sum_{n=1}^{\infty}\frac1{n^2+1}. $$ Since $\dfrac1{n^2+1}<\dfrac1{n^2}$ and $\sum1/n^2$ converges, the series converges (absolutely, since all terms are positive) by the Comparison Test.
Step 3 · Endpoint $x=-1/2$
$$ \sum_{n=1}^{\infty}\frac{2^n(-1/2)^n}{n^2+1} = \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+1}. $$ Since $\sum\left|\dfrac{(-1)^n}{n^2+1}\right|=\sum\dfrac1{n^2+1}$ converges (Step 2), this series converges absolutely as well.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1/2$ |
| At $x=1/2$ | Converges absolutely |
| At $x=-1/2$ | Converges absolutely |
| Interval of convergence | $[-1/2,1/2]$ |
Additional Notes
- This is the exponential-coefficient analogue of Problem 226: in both cases, a quadratic (or higher) polynomial in the denominator is exactly what's needed to secure absolute convergence at both endpoints, regardless of what drives the radius itself (here $2^n$, there a rational function).
- The factor $2^n$ in the numerator is entirely responsible for shrinking the radius from $1$ (which the bare series $\sum x^n/(n^2+1)$ would have) down to $1/2$; it plays no role at all in the endpoint analysis, since it exactly cancels against $(1/2)^n$ there.
- Since $1/(n^2+1)$ is the same majorant family used in Problems 194, 197, 198, and 208 for uniform-convergence arguments, this problem shows the identical comparison test doing double duty here to settle ordinary (rather than uniform) convergence at fixed points.