Question

How do you test the series `sum_(n=1)^oo n/(n^2+2)` for convergence?

Answer 1

This series diverges by the integral test.

Explanation:

Note that:

`x = 1/2 d/(dx) (x^2+2)`

So:

`int x/(x^2+2) dx = 1/2 ln abs(x^2+2) + C -> oo` as `x->oo`

So `sum_(n=1)^oo n/(n^2+2)` diverges by the integral test.

Alternatively, note that:

`sum_(n=1)^oo n/(n^2+2) = sum_(n=1)^oo 1/(n+2/n) >= sum_(n=1)^oo 1/(n+2) = sum_(n=3)^oo 1/n`

which diverges.