Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{x^{n^2}}{n!}. $$
Explanation of Method
The exponent $n^2$ grows much faster than the usual linear power in $n$, so the ratio test's exponent difference is $2n+1$ rather than $1$, producing a threshold at exactly $|x|=1$ rather than at some other radius. Since $n!$ in the denominator ultimately outpaces any fixed power once $|x|\le1$, the series converges (absolutely) on the closed interval $[-1,1]$ and diverges outside it.
- Apply the ratio test, tracking the exponent difference $(n+1)^2-n^2=2n+1$.
- Determine the threshold at $|x|=1$.
- Confirm convergence at both endpoints via direct comparison with $\sum1/n!$.
Step 1 · Ratio test
$$ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{(n+1)^2}/(n+1)!}{x^{n^2}/n!}\right| = \frac{|x|^{2n+1}}{n+1}. $$
Step 2 · Evaluate the limit
For $|x|<1$: $|x|^{2n+1}\to0$ faster than $n+1\to\infty$ grows, so the ratio $\to0<1$ — converges. For $|x|=1$: ratio $=1/(n+1)\to0<1$ — converges. For $|x|>1$: $|x|^{2n+1}/(n+1)\to\infty$ — diverges.
Step 3 · Confirm endpoints directly
At $x=\pm1$: $|x^{n^2}|=1$ for every $n$, so $\left|\dfrac{x^{n^2}}{n!}\right|=\dfrac1{n!}$, and $\sum1/n!$ converges (equals $e-1$ shifted, since it starts at $n=1$). So the series converges absolutely at both endpoints too, matching the ratio-test conclusion.
Summary Table
| Region | Behavior |
|---|---|
| $|x|\le1$ | Absolutely convergent |
| $|x|>1$ | Diverges |
Additional Notes
- Unlike an ordinary power series $\sum a_nx^n$, this series has "gaps" (only exponents that are perfect squares appear), a lacunary-series structure closely related to theta functions and Jacobi's classical $\vartheta$-function identities.
- The threshold falling exactly at $|x|=1$ (rather than some other radius) is a direct consequence of the $1/n!$ factor: since factorial growth dominates any fixed exponential rate, the only surviving question is whether $|x|\le1$ or $|x|>1$, with the quadratic exponent $n^2$ only affecting the rate of convergence/divergence, not the threshold itself.
- This is among the very few series in this batch whose entire boundary (both endpoints) is included in the region of absolute convergence — a consequence of the extremely fast-growing $n!$ denominator overwhelming even a marginal boundary value of $|x|=1$.
Result
Converges absolutely for $|x|\le1$; diverges for $|x|>1$: