Alternating nut

Relevant wiki: Maclaurin Series

Consider the Maclaurin series of $\ln(1+x)$ as follows:

$\begin{aligned} \ln (1+x) & = \sum_{\color{#3D99F6}n=1}^\infty \frac {(-1)^{n+1}x^n}n & \small \color{#3D99F6} \text{for }-1 < x \le 1 \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {(-1)^nx^{n+1}}{n+1} \\ \int \ln (1+x)\ dx & = \sum_{n=0}^\infty \frac {(-1)^nx^{n+2}}{(n+1)(n+2)} + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \int_0^1 \ln (1+x)\ dx & = \sum_{n=0}^\infty \frac {(-1)^n}{(n+1)(n+2)} = s \end{aligned}$

$\begin{aligned} \implies s & = \int_0^1 \ln (1+x)\ dx = (1+x)\ln(1+x) - x \bigg|_0^1 = 2\ln 2 - 1 \end{aligned}$

$\implies e^{s+1} = e^{2\ln 2} = e^{\ln 4} = \boxed{4}$

Using Taylor series, we can write $s(x)=\sum_{n=0}^\infty \frac{(-1)^nx^n}{(n+1)(n+2)}=\frac{(1+x)\log(1+x)-x}{x^2},$ which converges for $x\in[-1,1]$ . Then, $s=\log(4)-1$ , leading to the result.