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

Question

6025 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T19:33:30

It is convergent.

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

$sum_(n=1)^oosin(((2n-1)pi)/2)=-sum_(n=1)^oo(-1)^n/n$

This is an alternating series and $lim_(n->oo)1/n=a_n=0$ so it is convergent.

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