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