Prove that: sum from m,n=1 to infinity
Prove that: $$ \sum_{m,n=1}^{\infty} \frac1{(4n-1)^{2m}}=\frac14\ln2. $$
Explanation of Method
Summing the geometric series over $m$ for fixed $n$ collapses the double series to $\sum_n1/[8n(2n-1)]$; partial fractions reduce this to a combination $2\sum1/(2n-1)-\sum1/n$, which is exactly (twice) the regrouping that produces the alternating harmonic series $\ln2$ once the divergent parts are combined correctly.
- Sum the inner geometric series over $m$.
- Decompose via partial fractions into $2/(2n-1)-1/n$.
- Combine using the alternating-harmonic-series identity $2\sum\frac1{2n-1}-\sum\frac1n=2\ln2$.
Step 1 · Inner geometric sum
$$ \sum_{m=1}^{\infty}\frac1{(4n-1)^{2m}} = \frac{1/(4n-1)^2}{1-1/(4n-1)^2} = \frac1{(4n-1)^2-1} = \frac1{16n^2-8n} = \frac1{8n(2n-1)}. $$
Step 2 · Partial fractions
$\dfrac1{n(2n-1)}=\dfrac2{2n-1}-\dfrac1n$.
Step 3 · Combine via the alternating harmonic identity
Splitting the alternating harmonic series by parity: $$ \ln2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n = \sum_{n=1}^{\infty}\frac1{2n-1}-\sum_{n=1}^{\infty}\frac1{2n} = \sum_{n=1}^{\infty}\frac1{2n-1}-\frac12\sum_{n=1}^{\infty}\frac1n, $$ so $2\sum_{n=1}^{\infty}\frac1{2n-1}-\sum_{n=1}^{\infty}\frac1n=2\ln2$ (understood as the limit of matched partial sums, since neither piece converges alone). Therefore: $$ \sum_{n=1}^{\infty}\frac1{8n(2n-1)} = \frac18\left(2\sum\frac1{2n-1}-\sum\frac1n\right) = \frac18\cdot2\ln2=\frac{\ln2}4. $$
Summary Table
| Step | Result |
|---|---|
| Inner sum | $1/[8n(2n-1)]$ |
| Key identity | $2\sum1/(2n-1)-\sum1/n=2\ln2$ |
| Final value | $\ln2/4$ |
Additional Notes
- This problem is the "one power lower" companion to Problem 155's odd-power version $\sum1/(4n-1)^{2m+1}$: both share the same inner-sum denominator $8n(2n-1)$ up to the extra factor of $\frac1{4n-1}$, but this even-power version telescopes elementarily via the alternating harmonic identity, while the odd-power version requires the digamma function.
- The matched-partial-sum identity $2\sum_{n\le N}\frac1{2n-1}-\sum_{n\le N}\frac1n\to2\ln2$ is precisely the $N\to\infty$ limit used implicitly whenever divergent harmonic-type pieces are recombined into a convergent alternating series — the same technique underlying Problem 141's partial-fraction manipulation.
- Together with Problem 154's simpler identity $\sum1/(2n)^m=\ln2$, this problem shows how shifting the linear form inside the power (from $2n$ to $4n-1$) rescales the answer by a clean rational factor ($\ln2\to\ln2/4$) while the underlying mechanism (geometric sum, then partial fractions, then alternating harmonic series) stays the same.
Result
The identity is proven: