Answer
$\ln (2)$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as:
$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+....=\Sigma_{k=1}^{\infty}(-1)^{k+1} \dfrac{x^k}{k}$
Therefore, the Alternating Harmonic Series $\Sigma_{k=1}^{\infty}(-1)^{k+1} \dfrac{x^k}{k}$ for $f(1)$ can be written as:
$\Sigma_{k=1}^{\infty}(-1)^{k+1} \dfrac{x^k}{k}=1-\dfrac{1}{2}+\dfrac{1}{3}+.....\\=f(1)\\\ln (1+1)\\=\ln (2)$
