TITLE: Show $\sum_{n=1}^∞ \frac{(-1)^{n-1}}{n}=\ln(2)$
QUESTION [0 upvotes]: 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

REPLY [0 votes]: 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 $\ln(1 + 1)$.
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