Find the domain of definition of f(x)=sum from n=0
Find the domain of definition of $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}$.
Explanation of Method
Setting $t=\frac{x-1}{x+1}$, the series is exactly the Taylor series of $\arctan t$, which converges precisely for $|t|\le1$. Translating that condition back into $x$ via a single squaring step (rather than exhaustive case-work) pins down the domain directly.
- Recognize the series as $\arctan\left(\frac{x-1}{x+1}\right)$.
- Solve $\left|\frac{x-1}{x+1}\right|\le1$ algebraically.
- Confirm the boundary point is included, since the arctangent series converges at $|t|=1$.
Step 1 · Recognize the series
Since $\arctan t=\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}t^{2n+1}$ converges for $|t|\le1$: with $t=\dfrac{x-1}{x+1}$, $$ f(x) = \arctan\left(\frac{x-1}{x+1}\right). $$
Step 2 · Solve $|t|\le1$ algebraically
For $x\ne-1$: $$ \left|\frac{x-1}{x+1}\right|\le1 \iff |x-1|\le|x+1| \iff (x-1)^2\le(x+1)^2 \iff -4x\le0 \iff x\ge0 $$ (squaring is valid since both sides are non-negative).
Step 3 · Confirm the boundary
At $x=0$: $t=\dfrac{0-1}{0+1}=-1$, and the arctangent series converges at $t=-1$ (Leibniz test, since $1/(2n+1)\downarrow0$). So $x=0$ is included. Note $x=-1$ (where $t$ is undefined) satisfies $x<0$ and is automatically excluded.
Summary Table
| Property | Value |
|---|---|
| Closed form | $\arctan\!\left(\dfrac{x-1}{x+1}\right)$ |
| Convergence condition | $x\ge0$ |
| Domain of definition | $[0,\infty)$ |
Additional Notes
- The algebraic shortcut $|x-1|\le|x+1|\iff x\ge0$ (via squaring) is the same technique used in Problem 209 to pin down $\left|\frac{x}{x+1}\right|<1\iff x>-\tfrac12$ — both problems reduce a fractional-linear inequality to a single linear condition this way, avoiding tedious sign case-work.
- As $x\to\infty$: $\frac{x-1}{x+1}\to1^-$, so $f(x)\to\arctan(1)=\pi/4$; at $x=0$: $f(0)=\arctan(-1)=-\pi/4$. The function smoothly interpolates between these two values across its entire domain.
- This substitution $t=(x-1)/(x+1)$ is a Möbius transformation mapping $[0,\infty)$ onto $[-1,1)$ (with $x=\infty\mapsto t=1$), which is exactly what converts the arctangent series' natural domain $|t|\le1$ into the half-line domain found here.