0
$\begingroup$

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 $\log(2)$.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.