How do you find the sum of this series?

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1 Answer
Mar 12, 2018

S=1S=1

Explanation:

We want to find the sum

S=sum_(n=1)^oo4/((4n-3)(4n+1))S=n=14(4n3)(4n+1)

Using partial fractions

S=sum_(n=1)^oo1/(4n-3)-1/(4n+1)S=n=114n314n+1

The sum to k must be

S_k=sum_(n=1)^k1/(4n-3)-1/(4n+1)Sk=kn=114n314n+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