Expand f(x)=dfrac1x² in a series of powers of (x+1)

Expand $f(x)=\dfrac1{x^2}$ in a series of powers of $(x+1)$, and state where the expansion is valid.

Explanation of Method

Substituting $y=x+1$ turns $1/x^2$ into $1/(1-y)^2$, the derivative of the geometric series — so the coefficients are simply $n+1$ (as already computed in Problem 211). The region of convergence is the largest disk around $x=-1$ that avoids the singularity at $x=0$.

Substitute $y=x+1$, rewriting $1/x^2$ as $1/(1-y)^2$.
Apply the differentiated geometric series $1/(1-y)^2=\sum(n+1)y^n$.
Translate the convergence condition $|y|<1$ back to $x$.

Step 1 · Substitution

Let $y=x+1$, so $x=y-1$: $$ \frac1{x^2} = \frac1{(y-1)^2} = \frac1{(1-y)^2}. $$

Step 2 · Differentiated geometric series

From $\frac1{1-y}=\sum_{n=0}^{\infty}y^n$, differentiating gives $\frac1{(1-y)^2}=\sum_{n=0}^{\infty}(n+1)y^n$ (Problem 211). Substituting back $y=x+1$: $$ \frac1{x^2} = \sum_{n=0}^{\infty}(n+1)(x+1)^n. $$

Step 3 · Region of convergence

The series converges for $|y|<1$, i.e. $|x+1|<1$, i.e. $$ -2

Summary Table

Property	Value
Series	$\sum(n+1)(x+1)^n$
Radius of convergence	$R=1$
Region of validity	$-2

Additional Notes

The coefficients $(n+1)$ here are identical to Problem 223's power series $\sum(n+1)x^n=1/(1-x)^2$; this problem is exactly that series recentered at $x=-1$ instead of $x=0$, with the singularity relocated correspondingly from $x=1$ to $x=0$.
The radius of convergence, $1$, is forced entirely by the distance from the expansion center $x=-1$ to the function's only singularity at $x=0$ — a rule that applies to every rational-function expansion in this batch (compare Problem 246's general treatment of $1/(a-x)$).
Numerically at $x=-1/2$: the closed form gives $f(-1/2)=4$, matching $\sum(n+1)(1/2)^n=1/(1-1/2)^2=4$ — a useful sanity check of the geometric-derivative identity.

Result

$$ \frac1{x^2} = \sum_{n=0}^{\infty}(n+1)(x+1)^n, \quad \text{valid for } -2