Comparison and Limit Comparison Tests — Question 3

*Question 3 Consider $\displaystyle\sum_{n=1}^{\infty}\frac{\sin^2 n}{n^2}$ (radians).

1. Establish a pointwise trigonometric bound.

2. Compare with a standard $p$-series.

3. Classify the series and explain why no averaging argument is needed.

*Question 3 – Solution

For every real number $t$, $0\le\sin^2t\le1$. Hence for each positive integer $n$, $$0\le\frac{\sin^2n}{n^2}\le\frac1{n^2}.$$

The series $\sum1/n^2$ converges because it is a $p$-series with $p=2>1$.

The given nonnegative series is bounded term by term above by a convergent series, so it converges. No information about the average value or periodic distribution of $\sin^2n$ is required.