How do you determine the convergence or divergence of #Sigma 1/ncosnpi# from #[1,oo)#?

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Answer 1 · 2016-12-29T10:06:08

#sum_(n=1)^oo 1/ncospin = -ln2#

First, we note that #cosnpi=(-1)^n# so that the series can be written as:

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

This series satisfies Leibniz' criteria as:

#lim_n 1/n = 0#
#1/n > 1/(n+1)#

so the series is convergent.

To determine the sum, let's consider a geometric series of ratio #-alpha# with #0 < alpha < 1#.

#sum_(n=0)^oo (-1)^nalpha^n = 1/(1+alpha)#

This series is absolutely convergent and so it can be integrated term by term:

#int_0^x (dalpha)/(1+alpha) = sum_(n=0)^oo int_0^x (-1)^nalpha^ndalpha#

#ln|1+x| = sum_(n=0)^oo (-1)^nx^(n+1)/(n+1)=-sum_(n=0)^oo (-1)^(n+1)x^(n+1)/(n+1)=-sum_(n=1)^oo (-1)^n(x^n)/n#

This equation holds for every #x in (0,1)# and passing to the limit for #x->1# we have:

#ln2 = -sum_(n=1)^oo (-1)^n/n#