Expand (x-3)/((x+1)²) in a power series in x, using known
Expand $\dfrac{x-3}{(x+1)^2}$ in a power series in $x$, using known expansions, and indicate when it is valid.
Explanation of Method
Partial fractions split the function into $\frac1{1+x}$ and $\frac1{(1+x)^2}$, both of which have known Maclaurin series (the second being the derivative of the first, up to sign). Combining the two series termwise gives a single closed-form coefficient.
- Decompose via partial fractions.
- Expand each piece using $(1+x)^{-1}$ and $(1+x)^{-2}$.
- Combine into a single series and identify the region of convergence.
Step 1 · Partial fractions
Write $\dfrac{x-3}{(x+1)^2}=\dfrac A{x+1}+\dfrac B{(x+1)^2}$. Then $x-3=A(x+1)+B$; matching coefficients gives $A=1$, and $A+B=-3\implies B=-4$. So $$ \frac{x-3}{(x+1)^2} = \frac1{1+x}-\frac4{(1+x)^2}. $$
Step 2 · Expand each term
$$ \frac1{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n, \qquad \frac1{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n \quad (|x|<1), $$ the second obtained by differentiating the first and adjusting sign.
Step 3 · Combine
$$ \frac{x-3}{(x+1)^2} = \sum_{n=0}^{\infty}(-1)^n\bigl[1-4(n+1)\bigr]x^n = \sum_{n=0}^{\infty}(-1)^{n+1}(4n+3)x^n = -3+7x-11x^2+15x^3-\cdots $$ Both parent series converge for $|x|<1$ (singularity at $x=-1$), so the combined series is valid there too.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^{n+1}(4n+3)x^n$ |
| Coefficient pattern | Arithmetic: $3,7,11,15,\ldots$ (common difference $4$) |
| Region of validity | $|x|<1$ |
Additional Notes
- The coefficient magnitudes $3,7,11,15,\ldots$ form an arithmetic sequence with common difference $4$ — a direct consequence of combining a constant-magnitude series with a linearly-growing one via partial fractions.
- The radius of convergence, $1$, is dictated entirely by the double pole at $x=-1$; since both partial-fraction pieces share this same singularity, no cancellation occurs to extend the radius (unlike the cancellation examples in Problem 236).
- This partial-fractions-then-expand technique is the standard approach for any rational function with a repeated linear factor in the denominator, directly generalizing Problem 248's simpler $1/x^2$ case (which is the same computation without the extra linear numerator term).