Question

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

Answer 1

We can conclude that:

`sum_(n=0)^oo root(n)(n)/n^2`

is convergent by direct comparison with `sum_(n=0)^oo 1/n^2`.

Explanation:

Note that:

`root(n)(n)/n^2 = n^(1/n)/n^2 = (e^lnn)^(1/n)/n^2 = (e^(lnn/n))/n^2`

Now as:

`ln n < n => lnn/n < 1 => e^(lnn/n) < e`

we have:

`root(n)(n)/n^2 < e/n^2`

Based on the p-series test we know that:

`sum_(n=0)^oo e/n^2 = e*sum_(n=0)^oo 1/n^2`

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