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A student asks,

What is the sum of 2/(1*3*5) + 4/(3*5*7) + 6/(5*7*9) + ... ?

MathHelp replies,

Let's look at the general case. Let's find the sum from n=1 to infinity of

(An^2 + Bn + C)/((n-F)(n-G)(n-H))

The way to solve this is by using the method of partial fractions.

(An^2 + Bn + C)/((n-F)(n-G)(n-H)) =
(AF^2+BF+C)/((n-F)(F-G)(F-H)) + (AG^2+BG+C)/((G-F)(n-G)(G-H)) + (AH^2+BH+C)/((H-F)(H-G)(n-H))

Although it may not look simpler, we've reduced things to three simpler problems, which are:

Find the sum from n=1 to infinity of P/(n-F) + Q/(n-G) + R/(n-H), where

P = (AF^2+BF+C)/((F-G)(F-H)),
Q = (AG^2+BG+C)/((G-F)(G-H)), and
R = (AH^2+BH+C)/((H-F)(H-G))

TelescopeEdit

Before we move any further, I hope alarms are going off in your head to the effect that P/(n-F) doesn't converge! So the sum P/(n-F) + Q/(n-G) + R/(n-H) will converge only if P+Q+R=0, so let's add up P, Q, and R, and see what we get.

Amazingly, P+Q+R=A. Here's how you can see this clearly. First, express P, Q, R with a common denominator

P = (H-G)(AF^2+BF+C)/((G-H)(F-G)(H-F)),
Q = (F-H)(AG^2+BG+C)/((G-H)(F-G)(H-F)), and
R = (G-F)(AH^2+BH+C)/((G-H)(F-G)(H-F))

Now, the numerator of P+Q+R is

HAF^2+HBF+HC -GAF^2-GBF-GC +FAG^2+FBG+FC -HAG^2-HBG-HC +GAH^2+GBH+GC -FAH^2-FBH-FC

All the terms without an "A" cancel, leaving

(HAF^2 -GAF^2 +FAG^2 -HAG^2 +GAH^2 -FAH^2) / ((G-H)(F-G)(H-F))
A(HF^2 -GF^2 +FG^2 -HG^2 +GH^2 -FH^2) / ((G-H)(F-G)(H-F))
A((H-G)F^2 +(F-H)G^2 +(G-F)H^2) / ((G-H)(F-G)(H-F))
A((G-H)(H-F)(F-G)) / ((G-H)(F-G)(H-F))
A

So, as long as A is zero, as it is in our problem, P+Q+R=0, and our series "telescopes" nicely as long as F-G, G-H, and H-F are all integers.

Back to our problemEdit

In our problem, find sum, S, of

S = 2/(1*3*5) + 4/(3*5*7) + 6/(5*7*9) + ... .

We have the general term:

2n/((2n-1)(2n+1)(2n+3)) = (1/4)n/((n-1/2)(n+1/2)(n+3/2))

This is (An^2 + Bn + C)/((n-F)(n-G)(n-H)), where A=0, B=1/4, C=0, F=1/2, G=-1/2, and H=-3/2

Calculating P, Q, R, we get 1/16, 1/8, and -3/16, respectively, so

S = sum from n=1 to infinity of (1/16)/(n-1/2) + (2/16)/(n+1/2) - (3/16)/(n+3/2)
S = sum from n=1 to infinity of (1/8)/(2n-1) + (2/8)/(2n+1) - (3/8)/(2n+3)
S = (1/8)/1 + (2/8)/3 + (1/8)/3, so
S = 1/4

and everything else cancels

Partial telescopingEdit

[improvement needed: explain partial telescoping]

What if F, G, and H aren't exactly an integer apart? For example, what if we need to find the sum of

3/(1*3*6) + 5/(3*5*8) + 7/(5*7*10) + ... ?

. . . . . . (add the MathHelp explanation here, editors!)

Another approachEdit

jimmyrep, here's another way to look at this problem...

Start with the telescoping series

S = b1 + b2 + b3 + ... ,

where bn = an - an+1 , so if an converges to zero,

S = a1 = (a1-a2) + (a2-a3) + (a3-a4) + ...

I'm guessing you already have enough experience to know that if the denominators of a1, a2, a3, ... are 1*3, 3*5, 5*7, ..., then the denominators of the differences of successive elements b1, b2, b3, ... are 1*3*5, 3*5*7, 5*7*9, ...,

So this tells us what we need as the denominators of an. What about the numerators? For an to converge to zero, the numerators need to grow slower than n^2, so let's model the numerators as An+B. So a1, a2, a3, ..., an will be

(A+B)/(1*3), (2A+B)/(3*5), (3A+B)/(5*7), ..., (An+B)/((2n-1)(2n+1))

Now, finding the sequence b1 = (a1-a2), b2 = (a2-a3), ..., bn = (an-an+1), ... ,

(3A+4B)/(1*3*5), (5A+4B)/(3*5*7), ..., (2An+A+4B)/((2n-1)(2n+1)(2n+3))

Since the first two numerators of b1 and b2 are 2 and 4, we have

3A+4B=2, and
5A+4B=4.

Solving these, A=1 and B=-1/4, so a1, a2, a3, ..., an, ... are

3/4 / (1*3), 7/4 / (3*5), 11/4 / (5*7), ..., (n-1/4)/((2n-1)(2n+1))

So the sum b1 + b2 + b3 + ... is equal to a1 = (A+B)/(1*3) = (3/4)/3 = 1/4.

This method can be used to find the sum of any sequence similar to this one in which the numerators are in arithmetic sequence.

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