Question

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

Answer 1

`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`