Prove that: sum from m,n=1 to infinity
Prove that: $$ \sum_{m,n=1}^{\infty} \frac1{(4n-1)^{2m+1}}=\frac\pi8-\frac12\ln2. $$
Explanation of Method
Summing the geometric series over $m$ for fixed $n$ reduces the double series to $\sum_n\dfrac1{8n(2n-1)(4n-1)}$, whose partial-fraction decomposition splits into three individually divergent harmonic-type series. Rather than summing each piece separately, the combined partial sum is tracked via the digamma function's logarithmic asymptotic $\psi(x)\sim\ln x$, which makes the divergences cancel exactly and reveals the answer in terms of the classical special values $\psi(\tfrac12)$ and $\psi(\tfrac34)$.
- Sum the inner geometric series over $m$ and decompose via partial fractions.
- Express the combined partial sum using the digamma asymptotic $\sum_{n=0}^{N-1}\frac1{n+a}\approx\ln N-\psi(a)$.
- Substitute the known values $\psi(\tfrac12)=-\gamma-2\ln2$ and $\psi(\tfrac34)=-\gamma+\tfrac\pi2-3\ln2$.
Step 1 · Inner sum and partial fractions
With $x=1/(4n-1)$: $\sum_{m=1}^{\infty}x^{2m+1}=\dfrac{x^3}{1-x^2}=\dfrac1{(4n-1)[(4n-1)^2-1]}=\dfrac1{8n(2n-1)(4n-1)}$ (using $(4n-1)^2-1=8n(2n-1)$, as in Problem 156). Partial fractions give $\dfrac1{n(2n-1)(4n-1)}=\dfrac1n+\dfrac2{2n-1}-\dfrac8{4n-1}$.
Step 2 · Digamma asymptotic for the combined partial sum
Using $\sum_{n=0}^{N-1}\frac1{n+a}\approx\ln N-\psi(a)$ for each of the three pieces (reindexed to start at $n=0$): $$ T_N=\sum_{n=1}^N\frac1n+2\sum_{n=1}^N\frac1{2n-1}-8\sum_{n=1}^N\frac1{4n-1} \approx [\ln N+\gamma] + [\ln N-\psi(\tfrac12)] - [2\ln N-2\psi(\tfrac34)]. $$ The $\ln N$ terms cancel exactly, leaving $T=\gamma-\psi(\tfrac12)+2\psi(\tfrac34)$.
Step 3 · Substitute known digamma values
With $\psi(\tfrac12)=-\gamma-2\ln2$ and $\psi(\tfrac34)=-\gamma+\tfrac\pi2-3\ln2$: $$ T=\gamma-(-\gamma-2\ln2)+2\left(-\gamma+\frac\pi2-3\ln2\right) = \pi-4\ln2. $$ The final answer is $T/8$: $$ \sum_{m,n=1}^{\infty}\frac1{(4n-1)^{2m+1}} = \frac{\pi-4\ln2}8 = \frac\pi8-\frac12\ln2. $$
Summary Table
| Quantity | Value |
|---|---|
| $\psi(1/2)$ | $-\gamma-2\ln2$ |
| $\psi(3/4)$ | $-\gamma+\pi/2-3\ln2$ |
| Combined constant $T$ | $\pi-4\ln2$ |
| Final sum | $\pi/8-\tfrac12\ln2$ |
Additional Notes
- Unlike Problems 153, 154, 156, and 157, whose outer sums telescope cleanly via elementary partial fractions, this problem's three-term partial fraction decomposition ($1/n$, $2/(2n-1)$, $-8/(4n-1)$) does not telescope directly — each piece individually diverges, and only the digamma function's asymptotic expansion reveals the exact cancellation.
- The special values $\psi(\tfrac12)=-\gamma-2\ln2$ and $\psi(\tfrac34)=-\gamma+\tfrac\pi2-3\ln2$ are classical results (derivable from the reflection and duplication formulas for the digamma function), and their appearance here — producing both a $\pi$ term and a $\ln2$ term — mirrors the mixed arctangent/logarithm flavor anticipated by the problem's stated answer.
- This problem is the genuinely hard member of the double-series quartet (153–157): its answer combines both classical constants ($\pi$ from Problem 157's arctangent mechanism and $\ln2$ from Problems 154/156's logarithm mechanism) into a single expression, reflecting the odd-power structure $(4n-1)^{2m+1}$ that couples both effects.
Result
The identity is proven: