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

1 Answer
Dec 29, 2016

sum_(n=1)^oo 1/ncospin = -ln2

Explanation:

First, we note that cosnpi=(-1)^n so that the series can be written as:

sum_(n=1)^oo (-1)^n/n

This series satisfies Leibniz' criteria as:

lim_n 1/n = 0
1/n > 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)^oo (-1)^nalpha^n = 1/(1+alpha)

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

int_0^x (dalpha)/(1+alpha) = sum_(n=0)^oo int_0^x (-1)^nalpha^ndalpha

ln|1+x| = sum_(n=0)^oo (-1)^nx^(n+1)/(n+1)=-sum_(n=0)^oo (-1)^(n+1)x^(n+1)/(n+1)=-sum_(n=1)^oo (-1)^n(x^n)/n

This equation holds for every x in (0,1) and passing to the limit for x->1 we have:

ln2 = -sum_(n=1)^oo (-1)^n/n