1^()circ Find the radius of convergence depending
$1^\circ$ Find the radius of convergence depending on $a,\alpha$. $2^\circ$ For $\alpha=1$, $a\ge2$, examine the endpoints. $3^\circ$ For $\alpha=0$ and (briefly) $\alpha=-\tfrac13$, find the sum. $4^\circ$ Sum $\sum_{k=1}^{\infty}\dfrac{2^{2k}}{(2k)!}x^{2k}$.
Explanation of Method
The root test, combined with Stirling's approximation for $n!$, cleanly separates three regimes depending on whether $\alpha$ is less than, equal to, or greater than $1$. Endpoint behavior at $\alpha=1$ then follows from the same Stirling estimate, giving terms of order $1/\sqrt n$. The special-case sums for $\alpha=0$ and Part $4^\circ$ reduce directly to the exponential and hyperbolic-cosine series.
- Apply the root test with Stirling's approximation to find the radius in each regime of $\alpha$.
- Estimate the term size at $\alpha=1$'s endpoints and classify convergence.
- Evaluate the sums for $\alpha=0$ and the companion series in Part $4^\circ$ directly.
Part 1° · Radius of convergence
By the root test, using $(2^n+a^n)^{1/n}\to\max(2,|a|)$ and Stirling's $(n!)^{1/n}\sim n/e$: $$ |b_n|^{1/n} \sim \max(2,|a|)\cdot e\cdot n^{\alpha-1}. $$ $$ \boxed{R = \begin{cases} \infty & \alpha<1 \\ \dfrac1{e\max(2,|a|)} & \alpha=1 \\ 0 & \alpha>1 \end{cases}} $$
Part 2° · Endpoints for $\alpha=1$, $a\ge2$
Here $R=1/(ae)$. Using Stirling, $\dfrac{n^n}{n!e^n}\sim\dfrac1{\sqrt{2\pi n}}$, so at $x=\pm\frac1{ae}$ the terms have magnitude $\sim\dfrac C{\sqrt n}$. At $x=+\frac1{ae}$ all terms are positive, and $\sum1/\sqrt n$ diverges. At $x=-\frac1{ae}$ the terms alternate in sign with magnitude $\to0$ monotonically, so by the Leibniz test the series converges (conditionally).
Part 3° · Sum for $\alpha=0$
$$ \sum_{n=1}^{\infty}\frac{2^n+a^n}{n!}x^n = \sum_{n=1}^{\infty}\frac{(2x)^n}{n!}+\sum_{n=1}^{\infty}\frac{(ax)^n}{n!} = e^{2x}+e^{ax}-2. $$ For $\alpha=-\tfrac13$, the extra factor $n^{-n/3}$ does not correspond to any elementary function; the sum has no simple closed form.
Part 4° · The companion series
Since $\cosh t=\sum_{k=0}^{\infty}t^{2k}/(2k)!$, with $t=2x$: $$ \sum_{k=1}^{\infty}\frac{2^{2k}}{(2k)!}x^{2k} = \sum_{k=1}^{\infty}\frac{(2x)^{2k}}{(2k)!} = \cosh(2x)-1. $$
Summary Table
| Property | Value |
|---|---|
| Radius | $\infty$ ($\alpha<1$), $1/(e\max(2,|a|))$ ($\alpha=1$), $0$ ($\alpha>1$) |
| Endpoint at $+1/(ae)$ | diverges |
| Endpoint at $-1/(ae)$ | converges conditionally |
| Sum, $\alpha=0$ | $e^{2x}+e^{ax}-2$ |
| Part 4° sum | $\cosh(2x)-1$ |
Additional Notes
- The $\alpha=1$ boundary case is exactly where the "extra" $n^{\alpha n}=n^n$ growth in the numerator becomes precisely strong enough to compete with $n!$'s Stirling growth, producing the finite (rather than zero or infinite) radius.
- The asymmetry between the two endpoints — divergence on one side, conditional convergence on the other — mirrors the same $1/\sqrt n$-type behavior seen in Problem 312's structurally similar series, both governed by the same Stirling estimate.
- Part $4^\circ$'s series is a direct special case of the general even-power cosine-hyperbolic family built in Problem 253; here it's disguised by writing $2^{2k}$ instead of $(2x)^{2k}/x^{2k}$ explicitly.