Does anyone know how to find the exact sum of

$$ \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n} $$

I've only taken second semester calculus and don't see how to go about computing this sum. The only way that I know how to find the sum of an infinite series is if it is a geometric series.

Using WolframAlpha, I found that the sum for this series is $\log(2)$.


marked as duplicate by Hanul Jeon, Dominic Michaelis, Oleg567, Daniel Robert-Nicoud, Jyrki Lahtonen Nov 14 '13 at 6:52

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The Taylor series of the function $\ln{x}$ centered at $x = 1$ is given by

$$\sum\limits_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n} (x - 1)^n$$

Set $n = 2$ in this equation.

To prove this, note that if $f(x) = \ln{x}$, we have

\begin{align*} f'(x) &= \frac{1}{x} \\ f''(x) &= -\frac{1}{x^2} \\ f^{(3)}(x) &= \frac{2!}{x^3} \\ f^{(4)}(x) &= -\frac{3!}{x^4} \\ \end{align*}

and so on.


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