Show $\sum_{n=1}^∞ \frac{(-1)^{n-1}}{n}=\ln(2)$

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Taylor's theorem

a) If $x∈{0,1}$ and $n∈ℕ$, show that $$ \begin{vmatrix} ln(1+x)-(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+...+(-1)^{n-1}\frac{x^{n}}{n}) \\ \end{vmatrix}<\frac{x^{n+1}}{n+1} $$

b)Using the result of part "a" to Show $\sum_{n=1}^∞ \frac{(-1)^{n-1}}{n}=\ln(2)$ Note: If $(-1)^{0}=1$

I already proved part a, but I do not know how to do part b although I know it is simpler than a. If your could help me, I'll thank you

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One can check the convergence of the series by the alternating series test. Abel's theorem then tells us that the value is continuous from the left, meaning the value is $\ln(1 + 1)$.