# How do you determine whether the series is convergent or divergent given sum (sin^2(n))/(n*sqrt(n)) for n=1 to n=oo?

Mar 31, 2018

Converges by the Comparison Test.

#### Explanation:

On the interval $\left[1 , \infty\right] , {\sin}^{2} \left(n\right) < 1$ (not $\le 1 , n \ne \frac{\pi}{2} + 2 n \pi$ as we're dealing only with integers).

So, knowing that ${\sin}^{2} \left(n\right)$ is always less than one, we can remove it from the sequence ${a}_{n} = {\sin}^{2} \frac{n}{n \sqrt{n}}$ to create a new sequence

${b}_{n} = \frac{1}{n \sqrt{n}} = \frac{1}{n \left({n}^{\frac{1}{2}}\right)} = \frac{1}{n} ^ \left(\frac{3}{2}\right) \ge {a}_{n}$ for all $n .$ Removing a quantity that is always less than one from the numerator creates a new sequence that is always larger than the previous one.

So, now, we know ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{3}{2}\right)$ is convergent. It is a $p -$series in the form $\sum \frac{1}{n} ^ p$ where $p = \frac{3}{2} > 1$, so it must converge. We could prove this property by the Integral Test.

Then, since the larger series ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{3}{2}\right)$ converges, so must the smaller series ${\sum}_{n = 1}^{\infty} \frac{{\sin}^{2} n}{n \sqrt{n}}$.