Find the sum sum from n=1 to infinity
Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{2n(2n-1)}$.
Explanation of Method
Partial fractions split $\frac1{2n(2n-1)}$ into $\frac1{2n-1}-\frac1{2n}$; the resulting two pieces are recognized directly as (a factor of $x$ times) the $\arctan x$ series and (half of) the $\ln(1+x^2)$ series.
- Decompose via partial fractions.
- Identify each piece as a known series in disguise.
- Combine and verify against low-order terms and a numerical check.
Step 1 · Partial fractions
$$ \frac1{2n(2n-1)} = \frac1{2n-1}-\frac1{2n} \implies \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{2n(2n-1)} = \underbrace{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{2n-1}}_{A} \ - \ \underbrace{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{2n}}_{B}. $$
Step 2 · Identify each piece
Since $\arctan x=\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^{2n-1}}{2n-1}$, multiplying by $x$ gives $A=x\arctan x$. Since $\ln(1+t)=\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}t^n}n$ with $t=x^2$, we get $B=\dfrac12\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^{2n}}n=\dfrac12\ln(1+x^2)$.
Step 3 · Combine and verify
$$ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{2n(2n-1)} = x\arctan x - \frac12\ln(1+x^2). $$ Low-order check: expanding gives $x\arctan x\approx x^2-\frac{x^4}3$ and $\frac12\ln(1+x^2)\approx\frac{x^2}2-\frac{x^4}4$, so the difference $\approx\frac{x^2}2-\frac{x^4}{12}$, matching the original series' first two terms ($n=1$: $\frac{x^2}2$; $n=2$: $-\frac{x^4}{12}$) exactly. At $x=1$: $\frac\pi4-\frac12\ln2\approx0.785398-0.346574=0.438824$, consistent with the slowly-converging partial sums of the original series.
Summary Table
| Property | Value |
|---|---|
| Sum | $x\arctan x-\frac12\ln(1+x^2)$ |
| At $x=1$ | $\frac\pi4-\frac12\ln2\approx0.4388$ |
| Region of validity | $|x|\le1$ |
Additional Notes
- Both pieces here reuse series already built earlier in the collection: $\arctan x$ from Problem 267 and $\ln(1+x^2)$ (a substitution into Problem 293's logarithmic series with $x\to-x^2$, negated).
- This decomposition mirrors Problem 273's $(1+x^2)\arctan x$, both showing how multiplying a known series by $x$ or a low-degree polynomial, or dividing a coefficient via partial fractions, tends to produce combinations of elementary transcendental functions.
- The result stays finite and well-behaved up to $|x|=1$, since both $\arctan x$ and $\ln(1+x^2)$ are continuous there, unlike series with a genuine pole at the boundary.