The general term of the series is #=(-1)^(n+1)/(n(n+1))#
We perform a decomposition into partial fractions
#1/(n(n+1))=A/n+B/(n+1)#
#=(A(n+1)+Bn)/(n(n+1))#
So,
#1=A(n+1)+Bn#
When #n=0#, #=>#, #1=A#
When #n=-1#, #=>#, #1=-B#
Therefore,
#1/(n(n+1))=1/n-1/(n+1)#
#(-1)^(n+1)/(n(n+1))=(-1)^(n+1)/n-(-1)^(n+1)/(n+1)#
#sum_1^oo(-1)^(n+1)/(n(n+1))=sum_1 ^oo(-1)^(n+1)/n-sum_0^oo(-1)^(n+1)/(n+1)#
#ln(1+x)=sum_1^ ( oo)(-1)^(n+1)/n*x^n#
#sum_1^ ( oo)(-1)^(n+1)/n=ln2#
#sum_0^(oo)(-1)^(n+1)/(n+1)=sum_0^1(-1)^(n+1)/(n+1)-sum_1^oo(-1)^(n)x^(n+1)/(n+1)#
#sum_0^oo(-1)^(n)x^(n+1)/(n+1)=1-ln(1+x)#
#sum_0^ ( oo)(-1)^(n+1)/(n+1)=1-ln2#
#sum_1^oo(-1)^(n+1)/(n(n+1))=ln2-(1-ln2)=2ln2-1#
