How do you test the convergence of the series #cos(n) sin (pi/n)^2#?

Question

4185 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-16T17:35:11

#sum_(n=0)^oo cos(n)sin^2(pi/n)#

is absolutely convergent by direct comparison.

Use the inequalities:

#abs(cosx) <= 1#

#abs(sin x) <= abs (x)#

so that:

#abs(cos(n)sin^2(pi/n)) <= pi^2/n^2#

As:

#sum_(n=0)^oo pi^2/n^2 = pi^2 sum_(n=0)^oo 1/n^2# is convergent based on the p-series test, then also:

#sum_(n=0)^oo cos(n)sin^2(pi/n)#

is absolutely convergent by direct comparison.