Solve Summation by Parts for Sum[n/3^n]

[tarheelborn]

Homework Statement

Using summation by parts, find Sum[n/3^n].

Homework Equations

Sum[a_k*b_k] = s_n*b_(n+1) - Sum[s_k(b_(k+1)-b_k]

The Attempt at a Solution

Let a_k = 1/3^k and b_k = k. Then b_(k+1)-b_k = 1. But what is s_k? I know that it is 1/3 + 1/3^2 + 1/3^3 + ... but what is the general term? Thanks for your help.

 

[Tedjn]

What is the sequence s_n? I took a look at http://en.wikipedia.org/wiki/Summation_by_parts#Definition, and your equation doesn't seem to fit it exactly.

 

[tarheelborn]

S_n is the sequence of partial sums of a_k. My formula is from Goldberg's Methods of Real Analysis.

 

[gomunkul51]

sum(n*(1/3)^n)
very similar to a geometric series, after one differentiation.

sum(n*(1/3)^n) = (1/3)*sum(n*(1/3)^(n-1))

we know that the sum of a geometric series is 1/(1-q), here q=1/3.

sum = (1/3)*diff(1/(1-(1/3)))

sound familiar?

*it's sum found using the integration/differentiation by parts theorem.