How do you determine the convergence or divergence of #sum_(n=1)^(oo) (-1)^(n+1)/n#?
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The series is convergent and its sum is:
#sum_(n=1)^oo (-1)^(n+1)/n = ln2#
Leibniz's theorem states that a sufficient condition for the series with alternating signs:
#sum_(n=1)^oo (-1)^n a_n#
to be convergent is that:
(i) #a_(n+1) <= a_n#
(ii) #lim_(n->oo) a_n = 0#
Given: #a_n = 1/n# both conditions are satisfied so the series is convergent.
We can actually calculate its sum starting from the MacLaurin development of the function #ln(1+x)#
In fact:
#ln(1+x)|_(x=0) = 0#
#d/(dx) (ln(1+x)) = 1/(x+1) =(x+1)^(-1)#
#d^2/(dx^2) (ln(1+x)) = -(x+1)^(-2)#
#d^3/(dx^3) (ln(1+x)) = 2(x+1)^-3#
and we can easily conclude that:
#d^n/(dx^n) (ln(1+x)) =( -1)^(n-1)(n-1)! (x+1)^(-n)#
and for #x=0#
#d^n/(dx^n) (ln(1+x))|_(x=0) = ( -1)^(n-1)(n-1)! (1^(-n)) = ( -1)^(n-1)(n-1)!#
so that the MacLaurin series is:
#ln(1+x) = sum_(n=1)^oo (-1)^(n-1)(n-1)! x^n/(n!) = sum_(n=1)^oo (-1)^(n-1) x^n/n#
As #(-1)^(n-1) = (-1)^(n+1) # if we substitute #x=1# we have:
#ln2 = sum_(n=1)^oo (-1)^(n+1)/n#
