Macaroni and cheese

$\begin{aligned} S & = \sum_{j=2}^\infty \sum_{k=1}^\infty \frac {(-1)^j}{jk^j} \\ & = \sum_{k=1}^\infty \left(\frac 1k + \color{#3D99F6}{\sum_{j=1}^\infty \frac {(-1)^j}{jk^j}}\right) & \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \sum_{k=1}^\infty \left(\frac 1k \color{#3D99F6}{- \ln \left(1+\frac 1k \right)} \right) \\ & = \color{#3D99F6}{\gamma} \approx \boxed{0.577} & \small \color{#3D99F6}{\gamma = \text{Euler-Mascheroni constant}} \end{aligned}$

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References:

• Maclaurin series
• Euler-Mascheroni constant

The Euler- Mascheroni constant $\gamma$ fulfills $\displaystyle \gamma = \sum_{j = 2}^{\infty} (-1)^j \frac{\zeta(j)}{j} = \sum_{j = 2}^{\infty} (-1)^j \frac{\sum_{k = 1}^{\infty} \frac{1}{k^j}}{j} = \sum_{j = 2}^{\infty} \sum_{k = 1}^{\infty} \frac{(-1)^j}{jk^j} \approx 0.577215...$ There are a lot of forms for defining this constant... althought this result can be proved from other definition...

Note.- $\zeta(\cdot)$ is the Riemann-Zeta function