Series (Edexcel A-Level Further Mathematics): Revision Notes

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3.2.2 Method of Differences

Introduction

The method of differences is a powerful technique for summing series. It simplifies complex series by taking advantage of the telescoping property, where many terms cancel out. This method often involves breaking the general term of the series into partial fractions.

The basic idea is:

  1. Express the general term of the series in a form that can telescope (e.g., using partial fractions).
  2. Write out the expanded series.
  3. Cancel intermediate terms.
  4. Sum the remaining terms.

Partial Fractions and Telescoping Series

Partial Fractions

Partial fractions are used to rewrite a rational function as a sum of simpler fractions.

For example:

1r(r+1)=1r−1r+1\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}

This decomposition is key to creating a telescoping series.

Telescoping Series

A telescoping series has terms that cancel out in sequence.

For example:

∑r=1n(1r−1r+1)\sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right)

Expanding this sum:

(11−12)+(12−13)+⋯+(1n−1n+1)\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)

we see that all intermediate terms cancel, leaving:

1−1n+11 - \frac{1}{n+1}

Worked Examples

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Example 1:

Summing ∑r=1n1r(r+1)\sum_{r=1}^n \frac{1}{r(r+1)}

Step 1**: Decompose into partial fractions**

We start with:

1r(r+1)\frac{1}{r(r+1)}

Using partial fractions:

1r(r+1)=1r−1r+1\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}

Step 2**: Write the sum**

∑r=1n1r(r+1)=∑r=1n(1r−1r+1)\sum_{r=1}^n \frac{1}{r(r+1)} = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right)

Step 3**: Expand and telescope**

Expanding the series:

(11−12)+(12−13)+⋯+(1n−1n+1)\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)

Intermediate terms cancel, leaving:

1−1n+11 - \frac{1}{n+1}

Final Answer:

∑r=1n1r(r+1)=1−1n+1\sum_{r=1}^n \frac{1}{r(r+1)} = 1 - \frac{1}{n+1}
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Example 2:

Summing ∑r=1n1r(r+2)\sum_{r=1}^n \frac{1}{r(r+2)}

Step 1**: Decompose into partial fractions**

1r(r+2)\frac{1}{r(r+2)}

Using partial fractions:

1r(r+2)=12(1r−1r+2)\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)

Step 2**: Write the sum**

∑r=1n1r(r+2)=12∑r=1n(1r−1r+2)\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+2}\right)

Step 3**: Expand and telescope**

Expanding:

12[(11−13)+(12−14)+⋯+(1n−1n+2)]\frac{1}{2} \left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+2}\right)\right]

The series telescopes, leaving:

12(1+12−1n+1−1n+2)\frac{1}{2} \left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)

Final Answer:

∑r=1n1r(r+2)=12(1+12−1n+1−1n+2)\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)
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Example 3:

Summing ∑r=1nr+1r(r+2)\sum_{r=1}^n \frac{r+1}{r(r+2)}

Step 1**: Simplify the term**

r+1r(r+2)=rr(r+2)+1r(r+2)=1r+2+1r\frac{r+1}{r(r+2)} = \frac{r}{r(r+2)} + \frac{1}{r(r+2)} = \frac{1}{r+2} + \frac{1}{r}

Step 2**: Write the sum**

∑r=1nr+1r(r+2)=∑r=1n(1r+1r+2)\sum_{r=1}^n \frac{r+1}{r(r+2)} = \sum_{r=1}^n \left(\frac{1}{r} + \frac{1}{r+2}\right)

Step 3**: Expand and telescope**

Expanding:

(11+13)+(12+14)+⋯+(1n+1n+2)\left(\frac{1}{1} + \frac{1}{3}\right) + \left(\frac{1}{2} + \frac{1}{4}\right) + \dots + \left(\frac{1}{n} + \frac{1}{n+2}\right)

The series partially telescopes, leaving a combination of terms.

Final Answer:

The simplified form depends on the degree of nn but involves the remaining non-cancelled terms.

Note Summary

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Common Mistakes

  1. Incorrect partial fraction decomposition: Ensure the fraction is decomposed correctly.
  2. Failure to telescope properly: Not recognising which terms cancel.
  3. Omitting the last terms: Remember to include the non-cancelled terms at the end of the telescoping process.
  4. Arithmetic errors: Simplifying fractions or performing algebra incorrectly during expansion.
  5. Misapplying the method: Using the method on a series that does not telescope.
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Key Formulas

  1. Partial fraction decomposition:
1r(r+1)=1r−1r+1\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}1r(r+2)=12(1r−1r+2)\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)
  1. Telescoping series:
∑r=1n(1r−1r+1)=1−1n+1\sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1}
  1. Weighted terms: For 1r(r+k)\frac{1}{r(r+k)}, the partial fraction form is:
1r(r+k)=1k(1r−1r+k)\frac{1}{r(r+k)} = \frac{1}{k}\left(\frac{1}{r} - \frac{1}{r+k}\right)
  1. Series expansion: Expand the series to identify telescoping behaviour.
  2. Final sum: Include only the first and last non-cancelled terms.

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