# How do you determine the convergence or divergence of Sigma 1/ncosnpi from [1,oo)?

Dec 29, 2016

${\sum}_{n = 1}^{\infty} \frac{1}{n} \cos \pi n = - \ln 2$

#### Explanation:

First, we note that $\cos n \pi = {\left(- 1\right)}^{n}$ so that the series can be written as:

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

This series satisfies Leibniz' criteria as:

${\lim}_{n} \frac{1}{n} = 0$
$\frac{1}{n} > \frac{1}{n + 1}$

so the series is convergent.

To determine the sum, let's consider a geometric series of ratio $- \alpha$ with $0 < \alpha < 1$.

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\alpha}^{n} = \frac{1}{1 + \alpha}$

This series is absolutely convergent and so it can be integrated term by term:

${\int}_{0}^{x} \frac{\mathrm{da} l p h a}{1 + \alpha} = {\sum}_{n = 0}^{\infty} {\int}_{0}^{x} {\left(- 1\right)}^{n} {\alpha}^{n} \mathrm{da} l p h a$

$\ln | 1 + x | = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n + 1} / \left(n + 1\right) = - {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n + 1} {x}^{n + 1} / \left(n + 1\right) = - {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{n}}{n}$

This equation holds for every $x \in \left(0 , 1\right)$ and passing to the limit for $x \to 1$ we have:

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