Method of Differences

The method of differences (also called telescoping) is a technique for summing series where most terms cancel. It is used when the general term can be written as a difference $f(r) - f(r+1)$f(r)−f(r+1) (or similar), causing consecutive terms to cancel in the sum.

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The Key Idea

If we can write the general term as:

$u_r = f(r) - f(r+1)$ur​=f(r)−f(r+1)

then:

$\sum_{r=1}^{n} u_r = [f(1) - f(2)] + [f(2) - f(3)] + \cdots + [f(n) - f(n+1)] = f(1) - f(n+1)$∑r=1n​ur​=[f(1)−f(2)]+[f(2)−f(3)]+⋯+[f(n)−f(n+1)]=f(1)−f(n+1)

Most intermediate terms cancel — this is called telescoping.

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Using Partial Fractions

The most common application is to decompose a fraction using partial fractions, then telescope.

Worked Example 1: Find $\sum_{r=1}^{n} \frac{1}{r(r+1)}$∑r=1n​r(r+1)1​.

Partial fractions: $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$r(r+1)1​=r1​−r+11​.

$\sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$∑r=1n​(r1​−r+11​)=(1−21​)+(21​−31​)+⋯+(n1​−n+11​)

$= 1 - \frac{1}{n+1} = \frac{n}{n+1}$=1−n+11​=n+1n​

Worked Example 2: Find $\sum_{r=1}^{n} \frac{1}{r(r+2)}$∑r=1n​r(r+2)1​.

$\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$r(r+2)1​=21​(r1​−r+21​) (cover-up rule).

$\sum_{r=1}^{n} \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$∑r=1n​21​(r1​−r+21​)

Write out the first few and last few terms:

$= \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+2}\right)\right]$=21​[(1−31​)+(21​−41​)+(31​−51​)+⋯+(n1​−n+21​)]

After cancellation, the surviving terms are:

$= \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)$=21​(1+21​−n+11​−n+21​)=21​(23​−n+11​−n+21​)

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General Technique

1. Decompose the general term $u_r$ur​ into partial fractions (or another telescoping form).
2. Write out the first 2-3 and last 2-3 terms.
3. Cancel all intermediate terms.
4. Simplify the remaining terms.

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Non-Fraction Examples

The method of differences also works for non-fraction terms. If $u_r = f(r+1) - f(r)$ur​=f(r+1)−f(r):

Example 3: Given $u_r = (r+1)! - r!$ur​=(r+1)!−r!, find $\sum_{r=1}^{n} u_r$∑r=1n​ur​.

$\sum_{r=1}^{n} [(r+1)! - r!] = (n+1)! - 1! = (n+1)! - 1$∑r=1n​[(r+1)!−r!]=(n+1)!−1!=(n+1)!−1

Example 4: $\sum_{r=1}^{n} [r \cdot r! ] = \sum_{r=1}^{n} [(r+1)! - r!] = (n+1)! - 1$∑r=1n​[r⋅r!]=∑r=1n​[(r+1)!−r!]=(n+1)!−1 since $r \cdot r! = (r+1-1) \cdot r! = (r+1)! - r!$r⋅r!=(r+1−1)⋅r!=(r+1)!−r!.

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Practice Problems

1. Find $\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)}$∑r=1n​(2r−1)(2r+1)1​.
2. Find $\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)}$∑r=1n​r(r+1)(r+2)2​.
3. Show that $\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}$∑r=1n​r(r+1)1​=n+1n​ and deduce the value as $n \to \infty$n→∞.
4. Find $\sum_{r=1}^{20} \frac{1}{r(r+1)}$∑r=120​r(r+1)1​.

Solutions: