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

2 Answers
Oct 23, 2017

#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.

Oct 23, 2017

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