How do you find the sum of this series?

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Answer 1 · 2018-03-12T08:59:49

#S=1#

We want to find the sum

#S=sum_(n=1)^oo4/((4n-3)(4n+1))#

Using partial fractions

#S=sum_(n=1)^oo1/(4n-3)-1/(4n+1)#

The sum to k must be

#S_k=sum_(n=1)^k1/(4n-3)-1/(4n+1)#

Or writing out the terms

#S_k=(1-1/5)+(1/5-1/9)+...+(1/(4k-3)-1/(4k+1))#

Now we have a lot of cancellation

#S_k=1-1/(4k+1)#

Taking the limit as k goes to infinity

#S=lim_(k->oo)1-1/(4k+1)=1#