# Finding the value of $\sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n}$ [duplicate]

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 LahtonenNov 14 '13 at 6:52

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