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

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Answer 1 · 2017-10-23T09:57:03

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

is absolutely convergent.

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 · 2017-10-23T10:39:02

See below.

Considering that

#L=sum_(k=1)^oo 1/k^2 = pi^2/6# (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