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

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