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