1^()circ Find the radius of convergence depending
$1^\circ$ Find the radius of convergence depending on $\alpha$. $2^\circ$ For $\alpha=-1$, examine the endpoints. $3^\circ$ For $\alpha=0$, find the sum.
Explanation of Method
The root test with Stirling's approximation splits into three regimes exactly as in Problem 311, but here the dominant growth is the fixed base $3^n$ rather than an $a$-dependent one, giving a clean, parameter-free boundary case at $\alpha=-1$.
- Apply the root test to determine $R$ in each regime of $\alpha$.
- Estimate term size at the $\alpha=-1$ endpoints via Stirling's formula.
- Evaluate the sum directly for $\alpha=0$.
Part 1° · Radius of convergence
With $(3^n+n)^{1/n}\to3$ and $(n!)^{1/n}\sim n/e$: $$ |a_n|^{1/n} \sim \frac{3e}{n^{1+\alpha}}. $$ $$ \boxed{R = \begin{cases} \infty & \alpha>-1 \\ \dfrac1{3e} & \alpha=-1 \\ 0 & \alpha<-1 \end{cases}} $$
Part 2° · Endpoints for $\alpha=-1$
At $\alpha=-1$, $R=1/(3e)$, and using Stirling ($n^n/(n!e^n)\sim1/\sqrt{2\pi n}$), the general term at $x=\pm1/(3e)$ has magnitude $\sim1/\sqrt n$. At $x=+\frac1{3e}$: all terms positive, $\sum1/\sqrt n$ diverges. At $x=-\frac1{3e}$: terms alternate with magnitude $\to0$, so by the Leibniz test the series converges conditionally.
Part 3° · Sum for $\alpha=0$
$$ \sum_{n=1}^{\infty}\frac{3^n+n}{n!}x^n = \sum_{n=1}^{\infty}\frac{(3x)^n}{n!} + \sum_{n=1}^{\infty}\frac{nx^n}{n!} = (e^{3x}-1) + xe^x. $$
Summary Table
| Property | Value |
|---|---|
| Radius | $\infty\ (\alpha>-1)$, $1/(3e)\ (\alpha=-1)$, $0\ (\alpha<-1)$ |
| At $x=1/(3e)$ | diverges |
| At $x=-1/(3e)$ | converges conditionally |
| Sum, $\alpha=0$ | $e^{3x}+xe^x-1$ |
Additional Notes
- Because the dominant growth here is the fixed constant $3^n$ rather than a parameter-dependent $\max(2,|a|)^n$ as in Problem 311, the boundary value of $\alpha$ shifts from $1$ to $-1$ — the extra $n$-th root of $n!$ absorbs one more power of $n$ compared to the $n^{\alpha n}$ term already present in the denominator's exponent structure.
- The $1/\sqrt n$ endpoint-term behavior, and the resulting divergent-positive/conditionally-convergent-negative pattern, is structurally identical to Problem 311's $\alpha=1$ case — both are governed by the same Stirling-driven asymptotic.
- The $\alpha=0$ sum splits cleanly into a shifted exponential ($3^n/n!\to e^{3x}-1$) and a linear-weight exponential ($n/n!\to xe^x$, the same reindexing trick used in Problems 291–292 and 298).