Determine the radius of convergence for sum from n=1
Determine the radius of convergence for $\displaystyle\sum_{n=1}^{\infty}\left(\frac{na^{n^2}}{n+2}\right)^p\frac{x^n}{n!}$, where $a>0$, $p\in\mathbb R$.
Explanation of Method
The coefficient's dominant behavior is $a^{pn^2}/n!$: the polynomial factors $(n/(n+2))^p$ contribute nothing in the limit, so everything hinges on a tug-of-war between the doubly-exponential term $a^{pn^2}$ and the factorial. The sign of $p\ln a$ decides the winner.
- Simplify the ratio test, discarding the polynomial factors that tend to $1$.
- Isolate the dominant exponential factor $a^{p(2n+1)}$ against the factorial's contribution $1/(n+1)$.
- Determine $R$ by the sign of $p\ln a$.
Step 1 · Set up the ratio
With $c_n=\left(\dfrac n{n+2}\right)^p\dfrac{a^{pn^2}}{n!}$: $$ \left|\frac{c_{n+1}}{c_n}\right| = \left(\frac{(n+1)(n+2)}{n(n+3)}\right)^p \cdot a^{p[(n+1)^2-n^2]} \cdot \frac1{n+1} = \left(\frac{(n+1)(n+2)}{n(n+3)}\right)^p \cdot \frac{a^{p(2n+1)}}{n+1}. $$
Step 2 · Take the limit
Since $\dfrac{(n+1)(n+2)}{n(n+3)}\to1$, the polynomial factor contributes nothing in the limit. Writing $a^{p(2n+1)}=e^{p(2n+1)\ln a}$, the behavior is governed entirely by the sign of $p\ln a$: $$ \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| = \lim_{n\to\infty}\frac{a^{p(2n+1)}}{n+1}. $$
Step 3 · Determine the radius
If $p\ln a>0$ (i.e. $a>1,p>0$ or $0<a<1,p<0$): $a^{p(2n+1)}$ grows exponentially, dominating $1/(n+1)$, so the ratio $\to\infty$, giving $R=0$.
If $p\ln a<0$ (i.e. $a>1,p<0$ or $0<a<1,p>0$): $a^{p(2n+1)}\to0$ exponentially, so the ratio $\to0$, giving $R=\infty$.
If $p\ln a=0$ (i.e. $a=1$ or $p=0$): $a^{p(2n+1)}\equiv1$, so the ratio $=1/(n+1)\to0$, giving $R=\infty$.
Summary Table
| Condition | Radius $R$ |
|---|---|
| $p\ln a>0$ | $0$ |
| $p\ln a\le0$ | $\infty$ |
Additional Notes
- This is the sharpest coefficient-growth series in the batch: the doubly-exponential factor $a^{n^2}$ (when $p\ln a>0$) grows so fast that not even the factorial $n!$ can rescue convergence anywhere except $x=0$ — a stronger version of the growth seen in Problem 233's $(1+1/n)^{n^2}$.
- The special case $p=0$ collapses the series exactly to Problem 224's $\sum x^n/n!=e^x-1$, confirming $R=\infty$ directly, and the special case $a=1$ similarly reduces to a series dominated purely by $n!$ decay (since $\left(n/(n+2)\right)^p\to1$), also giving $R=\infty$.
- This all-or-nothing dichotomy ($R=0$ or $R=\infty$, with no finite intermediate value possible) is a hallmark of coefficients built from towers of exponentials — it never happens with the polynomial or single-exponential coefficients of Problems 223–232.