I remember solving this in highschool , but now I don't remember how to find sum of these kind of series .
I want to find the sum of the general series
Sum ∑∞n=1n.a−n=?
and ∑Nn=1n.a−n=?
Answer
Here is an approach that relies on the relationships (i) k=∑kℓ=1(1) and (ii) ∑Nk=ℓrk=rℓ−rN+11−r for |r|<1. Then, with r=3−1 we have
N∑k=1k3k=N∑k=13−kk∑ℓ=1(1)=N∑ℓ=1N∑k=ℓ3−k=N∑ℓ=13−ℓ−3−(N+1)1−1/3=32N∑ℓ=1(3−ℓ−3−(N+1))=32(3−1−3−(N+1)1−1/3)−N23−N=34−343−N−12N3−N
Note as N→∞ the sum of interest approaches 3/4.
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