Prove that: sum from m=2, n=1 to infinity frac1(2n)^m=ln2

Prove that: $$ \sum_{m=2,\,n=1}^{\infty} \frac1{(2n)^m}=\ln2. $$

Step 1 · Inner geometric sum

For fixed $n\ge1$, ratio $\frac1{2n}\le\frac12$: $$ \sum_{m=2}^{\infty}\frac1{(2n)^m} = \frac{1/(2n)^2}{1-1/(2n)} = \frac1{2n(2n-1)}. $$

Step 2 · Partial fractions

$\dfrac1{2n(2n-1)}=\dfrac1{2n-1}-\dfrac1{2n}$.

Step 3 · Recognize the alternating harmonic series

$$ \sum_{n=1}^{\infty}\left(\frac1{2n-1}-\frac1{2n}\right) = \left(1-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\cdots = 1-\frac12+\frac13-\frac14+\cdots = \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}k. $$ From the Taylor series $\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^k}k$ at $x=1$: this sum equals $\ln2$.

Summary Table

Additional Notes

• This is a striking instance of a double series over purely positive terms summing to the same constant $\ln2$ that governs the (signed) alternating harmonic series — the alternation here emerges organically from the partial-fraction decomposition, not from any explicit $(-1)^n$ factor in the original summand.
• This problem, together with Problems 156 and 157, forms a trio of related double series (all built from geometric sums over even powers of shifted linear forms in $n$) whose values are $\ln2$-, $\ln2/4$-, and $\pi/8$-type constants respectively, all traceable to classical Taylor series for $\ln(1+x)$ or $\arctan(x)$ at $x=1$.
• Since every term of the original double series is positive, Tonelli's theorem again guarantees the answer is independent of summation order, unlike the pathological example of Problem 150.