Find the sum sum from n=1 to infinity (n-1)/(n²(2n-1))
Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{n-1}{n^2(2n-1)}$.
Explanation of Method
Partial fractions split the term into $1/n+1/n^2-2/(2n-1)$; since $1/n$ and $2/(2n-1)$ individually diverge, they must be combined into a single convergent piece (using harmonic-number asymptotics) before adding the absolutely convergent $1/n^2$ piece.
- Decompose via partial fractions.
- Combine the two divergent pieces into one convergent limit using harmonic numbers.
- Add the convergent $\sum1/n^2$ piece and verify numerically.
Step 1 · Partial fractions
Solving $n-1=An(2n-1)+B(2n-1)+Cn^2$ gives $A=1$, $B=1$, $C=-2$: $$ \frac{n-1}{n^2(2n-1)} = \frac1n+\frac1{n^2}-\frac2{2n-1}. $$ (Check at $n=2$: $\frac12+\frac14-\frac23=\frac1{12}$, matching $\frac{2-1}{4\cdot3}=\frac1{12}$.)
Step 2 · Combine the divergent pieces
Since $\frac1n-\frac2{2n-1}=\dfrac{(2n-1)-2n}{n(2n-1)}=-\dfrac1{n(2n-1)}$, and using $\dfrac1{n(2n-1)}=\dfrac2{2n-1}-\dfrac1n$, the partial sums telescope via harmonic numbers: $$ \sum_{n=1}^N\left(\frac1n-\frac2{2n-1}\right) = 2H_N-2H_{2N} = -2(H_{2N}-H_N) \longrightarrow -2\ln2, $$ since $H_{2N}-H_N\to\ln2$ as $N\to\infty$.
Step 3 · Add the convergent piece and verify
$$ \sum_{n=1}^{\infty}\frac{n-1}{n^2(2n-1)} = \sum_{n=1}^{\infty}\frac1{n^2} + \sum_{n=1}^{\infty}\left(\frac1n-\frac2{2n-1}\right) = \frac{\pi^2}6-2\ln2 \approx 1.644934-1.386294 = 0.258640. $$ Numerical check: direct partial sums of the original series through $n=20$ give $\approx0.2346$, with an estimated tail of $\approx0.024$ (terms decay like $1/(2n^2)$), totaling $\approx0.258$ — consistent with this closed form.
Summary Table
| Property | Value |
|---|---|
| Decomposition | $1/n+1/n^2-2/(2n-1)$ |
| Sum | $\pi^2/6-2\ln2$ |
| Numerical value | $\approx0.2586$ |
Additional Notes
- The key identity $H_{2N}-H_N\to\ln2$ (a discrete analogue of $\int_N^{2N}dx/x=\ln2$) is what makes the divergent pieces combine into a finite limit — this same telescoping idea recurs whenever a partial-fraction decomposition splits a convergent series into individually divergent harmonic-type pieces.
- Equivalently, $\sum_{n=1}^\infty\frac1{n(2n-1)}=2\ln2$ is a clean standalone identity (verified independently via $\frac1{n(2n-1)}=\frac2{2n-1}-\frac1n$), and the original sum is just $\frac{\pi^2}6$ minus this value.
- This problem is a cautionary example: naively evaluating $\sum1/n$ and $\sum2/(2n-1)$ as separate divergent quantities and subtracting them is invalid; only the properly combined limit of partial sums gives the correct, finite answer.