Question

How do you determine the convergence or divergence of `Sigma sin(((2n-1)pi)/2)` from `[1,oo)`?

Answer 1

Not convergent.

Explanation:

`((2n-1)pi)/2=npi-pi/2`

so `sin(npi-pi/2)=-cos(npi) = -(-1)^n`

Finally

`sum sin(((2n-1)pi)/2) = -sum(-1)^n` which is not convergent