Find the domain of definition of f(x)=sum from n=1
Find the domain of definition of $\displaystyle f(x)=\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^p\left(\frac{\cos x-1}2\right)^n$.
Explanation of Method
Setting $y=\frac{\cos x-1}2=-\sin^2(x/2)$ confines $y$ to $[-1,0]$ for every real $x$. The coefficient series in $y$ has radius $1$ regardless of $p$, so the only question is what happens exactly at the boundary point $y=-1$ (i.e. $x=\pi+2k\pi$) — and there the outcome genuinely depends on the sign of $p$, since $(2n-1)!!/(2n)!!<1$ always, and raising a number below $1$ to a non-positive power sends it to $1$ or beyond.
- Substitute $y=\frac{\cos x-1}2=-\sin^2(x/2)\in[-1,0]$.
- Find the radius of convergence in $y$: it equals $1$ for every $p$.
- Resolve the boundary case $y=-1$ by the sign of $p$.
Step 1 · Substitution
Using $\cos x-1=-2\sin^2(x/2)$: $$ y = \frac{\cos x-1}2 = -\sin^2\!\left(\frac x2\right) \in [-1,0] \quad \text{for every real } x. $$
Step 2 · Radius of convergence in $y$
Let $c_n=\left[\dfrac{(2n-1)!!}{(2n)!!}\right]^p$. Since $\dfrac{(2n+1)!!/(2n+2)!!}{(2n-1)!!/(2n)!!}=\dfrac{2n+1}{2n+2}$: $$ \left|\frac{c_{n+1}}{c_n}\right| = \left(\frac{2n+1}{2n+2}\right)^p \longrightarrow 1^p=1. $$ So the radius is $1$ in $y$, for any real $p$. Since $|y|\le1$ always (Step 1), interior points $y\in(-1,0]$ converge absolutely automatically; only $y=-1$ (i.e. $x=\pi+2k\pi$) needs a separate check.
Step 3 · The boundary case $y=-1$
Since $(2n-1)!!/(2n)!!$ is a positive sequence strictly less than $1$ for every $n$ (and $\sim1/\sqrt{\pi n}\to0$), its $p$-th power $c_n$ behaves very differently depending on the sign of $p$: $$ c_n\to0 \text{ (monotonically) if } p>0, \qquad c_n\ge1 \text{ for all } n \text{ (in fact } c_n\to\infty\text{) if } p\le0. $$ If $p>0$: $c_n\downarrow0$, so by the Leibniz test the alternating series $\sum(-1)^nc_n$ converges at $y=-1$. If $p\le0$: $c_n\not\to0$ (indeed $c_n\ge1$ always), so the terms fail the necessary condition for convergence, and the series diverges at $y=-1$.
Summary Table
| Case | Domain |
|---|---|
| $p>0$ | $\mathbb R$ (all real $x$) |
| $p\le0$ | $\mathbb R\setminus\{\pi+2k\pi:k\in\mathbb Z\}$ |
Additional Notes
- This is a case where the radius of convergence alone ($R=1$, independent of $p$) does not settle the question — the entire domain answer hinges on a delicate boundary analysis, much like the endpoint tests of Problems 223–235, except here the "endpoint" recurs periodically in $x$ rather than appearing just twice.
- The double-factorial ratio $(2n-1)!!/(2n)!!\sim1/\sqrt{\pi n}$ is the same asymptotic that governs the central binomial coefficient $\binom{2n}n/4^n$; it decays just slowly enough that raising it to a positive power $p$ still gives a term sequence decreasing to $0$, letting the Leibniz test rescue convergence at $y=-1$.
- The excluded points $x=\pi+2k\pi$ for $p\le0$ are exactly where $\cos x=-1$, the unique points (mod $2\pi$) where $y$ reaches its extreme value $-1$ — everywhere else $|y|<1$ strictly, so convergence is automatic there regardless of $p$.