# How do you determine the convergence or divergence of sum_(n=1)^(oo) (-1)^(n+1)/n?

Jan 23, 2017

The series is convergent and its sum is:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n = \ln 2$

#### Explanation:

Leibniz's theorem states that a sufficient condition for the series with alternating signs:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$
to be convergent is that:

(i) ${a}_{n + 1} \le {a}_{n}$

(ii) ${\lim}_{n \to \infty} {a}_{n} = 0$

Given: ${a}_{n} = \frac{1}{n}$ both conditions are satisfied so the series is convergent.

We can actually calculate its sum starting from the MacLaurin development of the function $\ln \left(1 + x\right)$

In fact:

$\ln \left(1 + x\right) {|}_{x = 0} = 0$

$\frac{d}{\mathrm{dx}} \left(\ln \left(1 + x\right)\right) = \frac{1}{x + 1} = {\left(x + 1\right)}^{- 1}$

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left(\ln \left(1 + x\right)\right) = - {\left(x + 1\right)}^{- 2}$

${d}^{3} / \left({\mathrm{dx}}^{3}\right) \left(\ln \left(1 + x\right)\right) = 2 {\left(x + 1\right)}^{-} 3$

and we can easily conclude that:

d^n/(dx^n) (ln(1+x)) =( -1)^(n-1)(n-1)! (x+1)^(-n)

and for $x = 0$

d^n/(dx^n) (ln(1+x))|_(x=0) = ( -1)^(n-1)(n-1)! (1^(-n)) = ( -1)^(n-1)(n-1)!

so that the MacLaurin series is:

ln(1+x) = sum_(n=1)^oo (-1)^(n-1)(n-1)! x^n/(n!) = sum_(n=1)^oo (-1)^(n-1) x^n/n

As ${\left(- 1\right)}^{n - 1} = {\left(- 1\right)}^{n + 1}$ if we substitute $x = 1$ we have:

$\ln 2 = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n$