Question

How do you test the series `sum_(n=1)^(oo) sin^2n/n^2` for convergence?

Answer 1

Converges.

Explanation:

Recall the series

`sum_(n = 1)^oo 1/n^2`

By the p-series test, we know that it converges.

Because `sin^2n < 1` for all values of `n`, by the comparison test, we can confirm that this series converges as well.

Hopefully this helps!