Question

Why does the integral test not apply to `Sigma (-1)^n/n` from `[1,oo)`?

Answer 1

The integral test stipulates that for `suma_n` to apply, the terms of `a_n` must all be positive.

`(-1)^n/n` alternates between positive and negative terms.

You could use the integral test on `sum_(n=1)^oo1/n` since `1/n` is both positive and decreasing, which are the conditions for the integral test to apply.