1^()circ Find the domain of convergence, for x ≥0
$1^\circ$ Find the domain of convergence, for $x\ge0$, of $f(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{4n^2-1}\left(\dfrac{x-1}{x+1}\right)^{2n}$. $2^\circ$ Find a closed form for $f(x)$. $3^\circ$ Evaluate $\sum_{n=1}^{\infty}\dfrac{(-1)^n}{(4n^2-1)9^n}$.
Explanation of Method
Writing $u=\frac{x-1}{x+1}$, the series is $\sum\frac{(-1)^{n-1}}{4n^2-1}u^{2n}$, whose coefficients decay like $1/n^2$, so it converges absolutely for every $|u|\le1$ — in particular for all $x\ge0$. Partial fractions split $1/(4n^2-1)$ into two pieces that each relate to the arctangent series once the exponent is arranged as an odd power of $u$: one piece is exactly $u\arctan u$, and the other is $1-\arctan(u)/u$, up to sign.
- Determine the range of $u=\frac{x-1}{x+1}$ for $x\ge0$ and confirm absolute convergence there.
- Decompose via partial fractions, match each piece against the arctangent Taylor series, and combine.
- Evaluate at the $x$ giving $u=1/3$ for Part $3^\circ$.
Part 1° · Domain of convergence
Let $u=\dfrac{x-1}{x+1}$. As $x$ runs over $[0,\infty)$, $u$ increases monotonically from $u(0)=-1$ up to $u(\infty)=1$ (not attained), so $u\in[-1,1)$ and $u^2\in[0,1]$. Since $\left|\dfrac{(-1)^{n-1}}{4n^2-1}\right|\sim\dfrac1{4n^2}$, the series $\sum\dfrac1{4n^2-1}|u|^{2n}$ converges absolutely for every $|u|\le1$ by comparison with $\sum1/n^2$ — including at $x=0$ (where $u=-1$). Hence: $$ \boxed{\text{Domain: } x\in[0,\infty).} $$
Part 2° · Closed form
Partial fractions: $$ \frac1{4n^2-1}=\frac1{(2n-1)(2n+1)}=\frac12\left(\frac1{2n-1}-\frac1{2n+1}\right). $$ So $$ f(x) = \frac12\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac1{2n-1}-\frac1{2n+1}\right)u^{2n}. $$ For the first piece, using $\arctan u=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}u^{2n-1}$: $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}u^{2n} = u\arctan u. $$ For the second, from $\dfrac{\arctan u}u=\sum_{k=0}^\infty\dfrac{(-1)^k}{2k+1}u^{2k}=1-\dfrac{u^2}3+\dfrac{u^4}5-\cdots$, isolating the $k\ge1$ tail and flipping sign to match $(-1)^{n-1}$: $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n+1}u^{2n} = 1-\frac{\arctan u}{u}. $$ Combining: $$ f(x) = \frac12\left[u\arctan u - \left(1-\frac{\arctan u}u\right)\right] = \frac{(u^2+1)\arctan u}{2u} - \frac12. $$ (Verified numerically at $u=1/3$: closed form gives $0.036251$, matching direct term-by-term summation of $\sum\frac{(-1)^{n-1}}{4n^2-1}u^{2n}$ to 5 significant figures.) $$ \boxed{f(x) = \frac{(u^2+1)\arctan u}{2u} - \frac12,\qquad u=\frac{x-1}{x+1}.} $$
Part 3° · Evaluating the numerical series
The target series has $u^{2n}=9^{-n}$, so $u=1/3$, which corresponds to $x=2$ (since $\frac{2-1}{2+1}=\frac13$). Because the target has $(-1)^n$ rather than $(-1)^{n-1}$, it equals $-f(2)$: $$ f(2) = \frac{(1/9+1)\arctan(1/3)}{2/3}-\frac12 = \frac53\arctan\frac13-\frac12. $$ Therefore: $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{(4n^2-1)9^n} = -f(2) = \frac12-\frac53\arctan\frac13 \approx -0.036251. $$ (Verified against direct summation: $n=1$ term $=-1/27\approx-0.037037$, $n=2$ term $\approx+0.000823$, $n=3$ term $\approx-0.0000392$; partial sums converge to $\approx-0.036251$, matching to 6 significant figures.)
Summary Table
| Property | Value |
|---|---|
| Domain | $x\in[0,\infty)$ |
| Closed form | $f(x)=\dfrac{(u^2+1)\arctan u}{2u}-\dfrac12$, $u=\dfrac{x-1}{x+1}$ |
| Part 3° | $\dfrac12-\dfrac53\arctan\dfrac13\approx-0.036251$ |
Additional Notes
- The substitution $u=(x-1)/(x+1)$, mapping $x\in[0,\infty)$ onto $u\in[-1,1)$, is the same Möbius-type device used elsewhere in this collection to convert half-line domains into symmetric power-series intervals.
- Because the coefficients decay like $1/n^2$ — one full power faster than the borderline $1/n$ case — the series converges absolutely at both endpoints $u=\pm1$, unlike the conditionally-convergent endpoint behavior seen in Problems 309–312.
- The closed form's structure, $(u^2+1)\arctan(u)/(2u)-1/2$, is the "companion" identity to the simpler $u\arctan u$ piece appearing in Part 1 of the derivation — both come from splitting $1/(4n^2-1)$ by partial fractions and recognizing each half as a shift of the standard arctangent series.