The sum of the series 1/(1*2) - 1/(2*3) + 1/(3*4) - .... upto infinity is equal to?

1 Answer
Jun 17, 2017

The sum is =2ln2-1

Explanation:

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