3

As title, to calculate $$\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(2^m n+1)}$$

I tried to calculate $$\sum_{m=0}^{\infty}\frac{1}{2^m n+1}$$ but to no avail.

Then, as the addend looks similar to $$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^n}n=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$ , I tried: $$\ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}n=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$$ So, $$\frac{1}{2x} \ln\frac{1+x}{1-x} = \sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{2n-1}=1+\frac{x^2}{3}+\frac{x^4}{5}+\cdots$$ Again, $$\frac12\left(\frac{1}{2x} \ln\frac{1+x}{1-x} + \frac{1}{2xi} \ln\frac{1+xi}{1-xi}\right) = \sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{2n-1} + \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} x^{2n-2}}{2n-1}$$

$$=1+\frac{x^4}{5}+\frac{x^8}{9}+\cdots=\sum_{n=0}^{\infty} \dfrac{x^{4n}}{4n+1}$$ but still to no avail....

athos
  • 5,177

1 Answers1

6

\begin{align} \sum_{m=0}^\infty\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n(2^m n+1)} &=\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\int_0^1 x^{2^m n}\,dx \\&=\sum_{m=0}^\infty\int_0^1\log(1+x^{2^m})\,dx \\&=-\int_0^1\log(1-x)\,dx=1. \end{align} The second equality (i.e., the ability to move the summation over $n$ under the integral) is justified by the uniform convergence of the power series, thanks to Abel's theorem. The third equality follows from this.

metamorphy
  • 39,111