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

Oct 23, 2017

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / {\left(n + 1\right)}^{2}$

is absolutely convergent.

#### Explanation:

By direct comparison we can see that for $n \ge 1$:

$\frac{1}{n + 1} ^ 2 < \frac{1}{n} ^ 2$

As:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2$

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

${\sum}_{n = 1}^{\infty} \frac{1}{n + 1} ^ 2$

is convergent, and:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / {\left(n + 1\right)}^{2}$

is absolutely convergent.

Oct 23, 2017

See below.

#### Explanation:

Considering that

$L = {\sum}_{k = 1}^{\infty} \frac{1}{k} ^ 2 = {\pi}^{2} / 6$ (Basel problem)

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

we have

$L = {\sum}_{k = 1}^{\infty} \frac{1}{2 k} ^ 2 + {\sum}_{k = 1}^{\infty} \frac{1}{2 k - 1} ^ 2 = {S}_{p} + {S}_{i}$

but

${S}_{p} = \frac{1}{4} L$ then

${S}_{p} - {S}_{i} = \frac{1}{4} L - \left(L - \frac{1}{4} L\right) = - \frac{L}{2} = - {\pi}^{2} / 12$

then

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k} / {k}^{2} = - {\pi}^{2} / 12$ so converges