How do you find the sum of this series?
1 Answer
Mar 12, 2018
Explanation:
We want to find the sum
S=sum_(n=1)^oo4/((4n-3)(4n+1))S=∞∑n=14(4n−3)(4n+1)
Using partial fractions
S=sum_(n=1)^oo1/(4n-3)-1/(4n+1)S=∞∑n=114n−3−14n+1
The sum to k must be
S_k=sum_(n=1)^k1/(4n-3)-1/(4n+1)Sk=k∑n=114n−3−14n+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