Question

How do you determine if the series the converges conditionally, absolutely or diverges given `Sigma (-1)^(n+1)/(n+1)` from `[1,oo)`?

Answer 1

`sum_(n=1)^oo (-1)^(n+1)/(n+1)" "` is conditionally convergent.

Explanation:

First note that:

`sum_(n=1)^oo 1/(n+1) = 1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16+...`

`color(white)(sum_(n=1)^oo 1/(n+1)) > 1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+...`

`color(white)(sum_(n=1)^oo 1/(n+1)) = 1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16)+...`

`color(white)(sum_(n=1)^oo 1/(n+1)) = 1/2+1/2+1/2+1/2+..." "` diverges.

So our series is not absolutely convergent.

However, note that `1/(n+1)` is monotonically decreasing with limit `0` as `n->oo`.

Hence:

`sum_(n=1)^oo (-1)^(n+1)/(n+1)`

is conditionally convergent.