Let $A \; = \; \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{1}{k(2^nk+1)} \hspace{2cm} B \; = \; \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{1}{2k(2^{n+1}k+1)}$ Note that while $A$ is the sum of $\tfrac{1}{k(2^nk+1)}$ over all $n \ge 0$ and $k \ge 1$ , the term $B$ is the sum of $\tfrac{1}{k(2^nk+1)}$ over all $n \ge 0$ and even $k \ge 1$ . Now $\begin{aligned} B &= \tfrac12\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{k(2^nk+1)} \\ &= \tfrac12\left[\sum_{k=1}^\infty \sum_{n=0}^\infty \frac{1}{k(2^nk+1)} - \frac{1}{k(k+1)}\right] \\ &= \tfrac12(A-1) \end{aligned}$ But the sum we are interested in is $\sum_{k=1}^\infty \sum_{n=0}^\infty \frac{(-1)^{k-1}}{k(2^nk+1)} \; = \; A - 2B \; = \; \boxed{1}$

Note that the sum (say $S$ ) converges absolutely as $\dfrac{1}{k(1 + k2^n)} < \dfrac{1}{k^22^n}$ . Hence, we can interchange the order of summations. Interchanging and writing $\dfrac{1}{1+k2^n}$ as $\displaystyle \int_0^1 x^{k2^n} \, \text{d}x$ , we get the following :

$\begin{aligned} S &= \sum_{n = 0}^{\infty} \sum_{k = 1}^{\infty} \frac{(-1)^{k - 1}}{k}\int_0^1 x^{k2^n} \, \text{d}x \\ &= \sum_{n = 0}^{\infty} \int_0^1 \ln (1 + x^{2^n}) \, \text{d}x \quad (\text{by uniform convergence of power series})\\ &= \int_0^1 \ln \left( \frac{1}{1 - x} \right) \, \text{d}x \quad (\text{by Lebesgue's monotone convergence theorem}) \\ &= - \int_0^1 \ln x \, \text{d}x \\ &= \boxed{1} \end{aligned}$

Note: The following equation is also used to get the third equality : $\prod_{n = 0}^{N - 1} (1 + x^{2^n}) = \frac{1 - x^{2^N}}{1 - x}$

Let $\displaystyle \text{S} = \sum_{k=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^{k-1}}{k(k\cdot2^n+1)}$

Since $\text{S}$ is absolutely convergent, we can swap summations to get,

$\displaystyle \text{S} = \sum_{n=0}^{\infty} \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(k\cdot2^n+1)}$

$= \sum_{n=0}^{\infty} \sum_{k=1}^{\infty} \left[\dfrac{(-1)^{k-1}}{k} - \dfrac{(-1)^{k-1}}{k+\frac{1}{2^n}}\right]$

Splitting into even and odd terms, we have,

$= \sum_{n=0}^{\infty} \sum_{k=1}^{\infty}\left[\dfrac{(-1)^{k-1}}{k} - \left(\dfrac{1}{k+\frac{1}{2^n}} - \dfrac{1}{k+\frac{1}{2^{n+1}}} \right)\right]$

Converting to Digamma Function , we have,

$= \sum_{n=0}^{\infty} \left[\log 2 + \psi \left(\dfrac{1}{2^n} \right) - \psi \left( \dfrac{1} {2^{n+1}} \right) - 2^{n}\right]$

$= \lim_{N \to \infty}\sum_{n=0}^{N} \left[\log 2 + \psi \left(\dfrac{1}{2^n} \right) - \psi \left( \dfrac{1} {2^{n+1}} \right) - 2^{n}\right] \quad \quad (*)$

Clearly, $(*)$ telescopes. Evaluating it, we have,

$\text{S} = \boxed{1}$

Let S0 and S1 be the sums ∑k 1 k ∑ ∞ n=0 1 k2 n+1 with k running over all odd and all even positive integers, respectively, so that S = S1 −S0 = (S0 +S1)−2S0. In S1, we may write k = 2 to obtain S1 = ∞ ∑ =1 1 2 ∞ ∑ n=0 1 2

n+1 +1

1 2 (S0 +S1) + ∞ ∑ `=1 1

2 ( +1)

1 2 (S0 +S1) + 1 2 because the last sum telescopes; this immediately yields S = 1.