Question

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`?

Answer 1

Converges by the Comparison Test.

Explanation:

On the interval `[1, oo], sin^2(n)<1` (not `<=1, n ne pi/2+2npi` as we're dealing only with integers).

So, knowing that `sin^2(n)` is always less than one, we can remove it from the sequence `a_n=sin^2n/(nsqrtn)` to create a new sequence

`b_n=1/(nsqrtn)=1/(n(n^(1/2)))=1/n^(3/2)>=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)^oo1/n^(3/2)` is convergent. It is a `p-`series in the form `sum1/n^p` where `p=3/2>1`, so it must converge. We could prove this property by the Integral Test.

Then, since the larger series `sum_(n=1)^oo1/n^(3/2)` converges, so must the smaller series `sum_(n=1)^oo(sin^2n)/(nsqrtn)`.