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

In summary, the conversation is discussing how to use summation by parts to find the sum of the series n/3^n. One person suggests using a_k = 1/3^k and b_k = k, while the other questions the use of s_k and references a formula from Goldberg's Methods of Real Analysis. The final response suggests using differentiation by parts to solve the problem.
  • #1
tarheelborn
123
0

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.
 
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  • #3
S_n is the sequence of partial sums of a_k. My formula is from Goldberg's Methods of Real Analysis.
 
  • #4
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.
 

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

1. What is summation by parts?

Summation by parts is a mathematical technique used to evaluate infinite series. It involves breaking down a series into smaller parts and manipulating them to simplify the overall sum.

2. How do you solve summation by parts for Sum[n/3^n]?

To solve this summation, we can use the following formula: Sum[f(n)g(n)] = f(n)g(n) - Sum[f(n+1)(Δg(n))]. In this case, f(n) = n and g(n) = 1/3^n. So, we can rewrite the summation as n(1/3^n) - Sum[(n+1)(Δ1/3^n)].

3. Can you provide an example of solving Sum[n/3^n] by parts?

Sure, let's say we want to evaluate the summation from n=1 to 5. We can use the formula mentioned above and plug in the values for f(n) and g(n):

Sum[n/3^n] = n(1/3^n) - Sum[(n+1)(Δ1/3^n)]

= (1/3) - [(2/3)(1/3) + (3/3)(1/9) + (4/3)(1/27) + (5/3)(1/81)]

= (1/3) - (2/9 + 1/9 + 4/81 + 5/243)

= (1/3) - (8/81)

= 19/81

4. What is the purpose of using summation by parts?

Summation by parts allows us to evaluate infinite series that cannot be solved using traditional methods. It also helps us to simplify and manipulate complex series to find their sums.

5. Are there any limitations to using summation by parts?

Summation by parts may not work for all types of series. It is most effective for series that have a common ratio between terms or can be rewritten in a way that allows us to use the formula. It also requires some knowledge of calculus and algebra to apply the formula correctly.

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