How do you test the series Sigma rootn(n)/n^2 from n is [1,oo) for convergence?

Jan 16, 2017

We can conclude that:

${\sum}_{n = 0}^{\infty} \frac{\sqrt[n]{n}}{n} ^ 2$

is convergent by direct comparison with ${\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2$.

Explanation:

Note that:

$\frac{\sqrt[n]{n}}{n} ^ 2 = {n}^{\frac{1}{n}} / {n}^{2} = {\left({e}^{\ln} n\right)}^{\frac{1}{n}} / {n}^{2} = \frac{{e}^{\ln \frac{n}{n}}}{n} ^ 2$

Now as:

$\ln n < n \implies \ln \frac{n}{n} < 1 \implies {e}^{\ln \frac{n}{n}} < e$

we have:

$\frac{\sqrt[n]{n}}{n} ^ 2 < \frac{e}{n} ^ 2$

Based on the p-series test we know that:

${\sum}_{n = 0}^{\infty} \frac{e}{n} ^ 2 = e \cdot {\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2$

is convergent, so that by direct comparison we can conclude that our series is convergent.