Convergence and Divergence of Series — Question 3

Question 3 – Solution

Step 1: Audit the obvious comparison.

The terms satisfy $0\le\sin^2n/n\le1/n$ and approach zero. However, an upper comparison with a divergent series is logically inconclusive: a smaller positive series may converge or diverge.

Step 2: Separate the average and oscillatory parts.

Using $\sin^2n=(1-\cos2n)/2$ gives $$s_N=\frac12\sum_{n=1}^N\frac1n-\frac12\sum_{n=1}^N\frac{\cos(2n)}n.$$

Step 3: Control the oscillatory series.

To apply Dirichlet’s Test to $\sum\cos(2n)/n$, verify both hypotheses. First, the partial sums of $\cos(2n)$ are bounded. Indeed, $$\sum_{n=1}^N e^{2in}=e^{2i}\frac{1-e^{2iN}}{1-e^{2i}},$$ whose magnitude is at most $2/|1-e^{2i}|$; taking real parts preserves boundedness. Second, $1/n$ is positive, decreasing, and tends to zero. Dirichlet’s Test therefore shows that $\sum\cos(2n)/n$ converges.

Step 4: Combine the two parts.

The first part of $s_N$ is one-half of the harmonic partial sum and tends to $+\infty$, while the oscillatory part approaches a finite limit. Subtracting a bounded quantity cannot stop the harmonic growth. Hence $$\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{\sin^2n}{n}\text{ diverges}.}$$ The needed extra structure is that $\sin^2n$ has positive average $1/2$, not merely that it is bounded.