Question

How do you determine if the series the converges conditionally, absolutely or diverges given `Sigma (-1)^(n+1)/(n+1)^2` from `[1,oo)`?

Answer 1

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

is absolutely convergent.

Explanation:

By direct comparison we can see that for `n>=1`:

`1/(n+1)^2 < 1/n^2`

As:

`sum_(n=1)^oo 1/n^2`

is convergent based on the `p`-series test, then also:

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

is convergent, and:

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

is absolutely convergent.

Answer 2

See below.

Explanation:

Considering that

`L=sum_(k=1)^oo 1/k^2 = pi^2/6` (Basel problem)

https://en.wikipedia.org/wiki/Basel_problem

we have

`L = sum_(k=1)^oo 1/(2k)^2 + sum_(k=1)^oo 1/(2k-1)^2 = S_p+S_i`

but

`S_p = 1/4 L` then

`S_p-S_i = 1/4L-(L-1/4L)= -L/2 = -pi^2/12`

then

`sum_(k=1)^oo(-1)^k/k^2 = -pi^2/12` so converges