Prove that: sum from m,n=1 to infinity
Prove that: $$ \sum_{m,n=1}^{\infty} \frac1{(4n-2)^{2m}}=\frac\pi8. $$
Explanation of Method
Writing $4n-2=2(2n-1)$ and summing the geometric series over $m$ first reduces the double series to $\sum_n1/[(4n-1)(4n-3)]$, whose partial-fraction telescoping regroups exactly into the Gregory–Leibniz series $1-\frac13+\frac15-\frac17+\cdots$ for $\arctan(1)=\pi/4$.
- Simplify $(4n-2)^{2m}=4^m(2n-1)^{2m}$ and sum the inner geometric series.
- Decompose via partial fractions into a telescoping form.
- Recognize the Gregory–Leibniz series for $\pi/4$.
Step 1 · Inner geometric sum
With $(4n-2)^{2m}=4^m(2n-1)^{2m}$: $$ \sum_{m=1}^{\infty}\frac1{[4(2n-1)^2]^m} = \frac1{4(2n-1)^2-1} = \frac1{16n^2-16n+3} = \frac1{(4n-1)(4n-3)}. $$
Step 2 · Partial fractions
$\dfrac1{(4n-1)(4n-3)}=\dfrac12\left(\dfrac1{4n-3}-\dfrac1{4n-1}\right)$.
Step 3 · Recognize the Gregory–Leibniz series
$$ \sum_{n=1}^{\infty}\frac12\left(\frac1{4n-3}-\frac1{4n-1}\right) = \frac12\left(1-\frac13+\frac15-\frac17+\cdots\right) = \frac12\arctan(1) = \frac12\cdot\frac\pi4 = \frac\pi8. $$
Summary Table
| Step | Result |
|---|---|
| Inner sum | $1/[(4n-1)(4n-3)]$ |
| Regrouped series | $1-\frac13+\frac15-\cdots=\pi/4$ |
| Final value | $\pi/8$ |
Additional Notes
- This is the $\pi$-flavored counterpart to Problem 154's $\ln2$ result: replacing the linear form $2n$ with $4n-2$ (equivalently doubling the "step" between consecutive terms) shifts the emergent series from the alternating harmonic series to the Gregory–Leibniz series, swapping $\ln2$ for $\pi/4$.
- The factorization $4(2n-1)^2-1=(4n-1)(4n-3)$ mirrors the analogous factorization $(4n-1)^2-1=8n(2n-1)$ used in Problems 155 and 156, both instances of a difference-of-squares trick applied to a shifted linear form.
- Since $4n-3$ and $4n-1$ are exactly the odd numbers congruent to $1$ and $3\pmod4$, this telescoping recovers the classical Leibniz series by picking out every other pair of odd reciprocals with alternating sign — the same series appearing in Problem 138's numerical estimate discussion of oscillatory sums.
Result
The identity is proven: