Find the domain of definition of f(x)=sum from n=1
Find the domain of definition of $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{n!}{a^{n^2}}\tan^nx$, $a>1$.
Explanation of Method
As a series in $t=\tan x$, the coefficients $n!/a^{n^2}$ vanish so fast (a factorial divided by a doubly-exponential) that the root test gives an infinite radius in $t$ — the series converges for every real value $\tan x$ can take. So the only restriction on the domain comes from where $\tan x$ itself is defined.
- Apply the root test to the series in $t=\tan x$.
- Show the radius (in $t$) is infinite, using Stirling's approximation.
- Conclude the domain is restricted only by the domain of $\tan x$.
Step 1 · Root test in $t=\tan x$
With $c_n=n!/a^{n^2}$: $$ \sqrt[n]{|c_n|} = \frac{(n!)^{1/n}}{a^n}. $$
Step 2 · Evaluate via Stirling's approximation
Since $n!\sim\sqrt{2\pi n}\,(n/e)^n$: $(n!)^{1/n}\sim n/e$. So $$ \sqrt[n]{|c_n|} \sim \frac{n}{ea^n}. $$ Since $a>1$, $a^n$ grows exponentially while $n$ grows only linearly, so $\sqrt[n]{|c_n|}\to0$. The radius (in $t$) is therefore infinite: the series in $t$ converges for every real $t$.
Step 3 · Domain in $x$
Since the series converges for every finite value of $\tan x$, the only obstruction to $f(x)$ being defined is where $\tan x$ itself is undefined, namely at $x=\dfrac{\pi}2+k\pi$, $k\in\mathbb Z$. So $$ \text{Domain: } x\in\mathbb R\setminus\left\{\frac{\pi}2+k\pi : k\in\mathbb Z\right\}. $$
Summary Table
| Property | Value |
|---|---|
| Radius (in $t=\tan x$) | $\infty$ |
| Domain (in $x$) | $\mathbb R\setminus\{\pi/2+k\pi\}$ |
Additional Notes
- This is the most extreme instance of factorial-versus-exponential dominance in this batch: whereas Problem 235's $a^{n^2}/n!$ can make the radius collapse to $0$ when the doubly-exponential term is in the numerator, here it sits in the denominator, and combined with $n!$ in the numerator the two still lose decisively to $a^{n^2}$ — giving infinite radius instead.
- Since the domain restriction comes entirely from the elementary function $\tan x$ rather than from the series itself, this problem illustrates a case where the "hard part" of a convergence analysis (the root/ratio test) turns out to impose no restriction at all — a pleasant contrast to most of the other domain problems in this collection (Problems 206–208, 239).
- The condition $a>1$ is essential: if $0<a<1$, then $a^{n^2}\to0$, so $n!/a^{n^2}=n!\cdot a^{-n^2}\to\infty$ even faster, and the series would instead have radius $0$ in $t$, drastically shrinking the domain.