Expand ln²(1-x) in a power series in x
Expand $\ln^2(1-x)$ in a power series in $x$.
Explanation of Method
Squaring the $\ln(1-x)$ series requires a Cauchy product; the resulting double sum telescopes into a clean closed form involving the harmonic numbers $H_n=\sum_{j=1}^n1/j$.
- Write $\ln(1-x)=-\sum_{n\ge1}x^n/n$ and set up the Cauchy product for its square.
- Simplify the coefficient $c_k=\sum_{n=1}^{k-1}\frac1{n(k-n)}$ using partial fractions.
- Identify the harmonic-number closed form and check numerically.
Step 1 · Set up the Cauchy product
$$ \ln^2(1-x) = \left(\sum_{n=1}^{\infty}\frac{x^n}n\right)^2 = \sum_{k=2}^{\infty}c_kx^k, \qquad c_k=\sum_{n=1}^{k-1}\frac1{n(k-n)}. $$
Step 2 · Simplify via partial fractions
Since $\dfrac1{n(k-n)}=\dfrac1k\left(\dfrac1n+\dfrac1{k-n}\right)$, and $\sum_{n=1}^{k-1}\frac1{k-n}=\sum_{n=1}^{k-1}\frac1n$ by symmetry: $$ c_k = \frac2k\sum_{n=1}^{k-1}\frac1n = \frac{2H_{k-1}}k. $$
Step 3 · Write the series and verify
$$ \ln^2(1-x) = \sum_{k=2}^{\infty}\frac{2H_{k-1}}kx^k = x^2+x^3+\frac{11x^4}{12}+\frac{5x^5}6+\cdots $$ Direct low-order check: $c_2=2H_1/2=1$; $c_3=2H_2/3=2(3/2)/3=1$; $c_4=2H_3/4=2(11/6)/4=11/12$ — all match the coefficients obtained by directly multiplying the series through $x^4$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum_{k=2}^{\infty}\frac{2H_{k-1}}kx^k$ |
| $H_n$ | $n$-th harmonic number |
| Region of validity | $|x|<1$ |
Additional Notes
- The series starts at $k=2$ (no constant or linear term), since $\ln(1-x)\to0$ as $x\to0$, and squaring a series with no constant term always pushes the lowest surviving power up by (at least) double the original lowest power.
- The coefficients grow like $\frac{2\ln k}k$ for large $k$ (since $H_{k-1}\sim\ln k$), which is exactly the borderline decay rate expected at the radius-of-convergence boundary $|x|=1$ — consistent with $\ln^2(1-x)\to\infty$ as $x\to1^-$.
- This harmonic-number coefficient pattern is closely related to the polylogarithm identity $\ln^2(1-x)=2\sum_{k=1}^\infty\frac{H_k}{k+1}x^{k+1}$, and is a standard building block for evaluating Euler sums.