# Does the series sum_{n=1} ^oo (2)/(n^2sqrtn) converge or diverge?

Jan 15, 2018

Converge

#### Explanation:

Use the comparison test.

We know that (from the Riemann-zeta function):

${\sum}_{n = 1}^{\infty} \frac{2}{n} ^ 2 = 2 {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = 2 \zeta \left(2\right) = 2 \cdot {\pi}^{2} / 6 = {\pi}^{2} / 3$

So

${\sum}_{n = 1}^{\infty} \frac{2}{n} ^ 2$ converges.

We also have that for all values of $n$:

$\frac{2}{{n}^{2} \sqrt{n}} < \frac{2}{n} ^ 2$ therefore:

sum_(n=1)^oo2/(n^2sqrt(n) must also converge by the comparison test.