Porisma

Worked solution

Reindex a rational series by two methods

Given ∑r=2n3r2+1(r2−1)3=12[98−1n3−1(n+1)3]\sum_{r=2}^n\dfrac{3r^2+1}{(r^2-1)^3}=\dfrac12\left[\dfrac98-\dfrac1{n^3}-\dfrac1{(n+1)^3}\right], find ∑r=3n3(r−1)2+1[(r−1)2−1]3\sum_{r=3}^n\dfrac{3(r-1)^2+1}{[(r-1)^2-1]^3} in terms of nn. Hence find its infinite sum.

[2009/SAJC/II/3(modified)]

CJC · Discussion · Question 12

Answer

Finite sum 12[9/8−1/(n−1)3−1/n3]\tfrac12[9/8-1/(n-1)^3-1/n^3]; infinite sum 9/169/16.

Method 1: change the index explicitly

Put s=r−1s=r-1. The lower bound becomes s=2s=2 and the upper bound s=n−1s=n-1. Applying the identity with upper bound n−1n-1 gives 12[9/8−1/(n−1)3−1/n3]\tfrac12[9/8-1/(n-1)^3-1/n^3].

How did this part go?

Method 2: compare the first and last summands

The new sum begins with [3(2)2+1]/(22−1)3[3(2)^2+1]/(2^2-1)^3, followed by [3(3)2+1]/(32−1)3[3(3)^2+1]/(3^2-1)^3, and ends at the term with n−1n-1 in place of the original index. Thus it is exactly the original sum through n−1n-1, giving the same expression.

How did this part go?

Take the infinite limit

Both reciprocal cubes tend to zero, so the infinite sum is 12(9/8)=9/16\tfrac12(9/8)=9/16.

How did this part go?

Common mistakes

The upper index becomes n−1, not n.

Read the hint