The method of differences, expressing a general term as a difference of consecutive terms (often via partial fractions), summing by cancellation, and finding the sum to infinity where it exists.

What this dot point is asking

OCR wants you to use the method of differences: express the general term of a series as a difference of consecutive (or near-consecutive) terms of some sequence, usually by splitting into partial fractions, sum the series by cancelling the telescoping middle terms, write the result in terms of $n$, and find the sum to infinity by letting $n \to \infty$ when the remaining terms have a finite limit.

The telescoping idea

A telescoping sum is one where each term cancels part of the next, like the sections of a collapsible telescope. If the general term is $u_r = f(r) - f(r+1)$, then the sum from $1$ to $n$ collapses to just $f(1) - f(n+1)$.

Setting up with partial fractions

Most exam terms are rational expressions that must first be split into partial fractions to reveal the difference structure. The denominators of the partial fractions differ by a fixed shift, which is exactly what telescopes.

Identifying the surviving terms

The safe method is to write out the first two or three and the last two or three terms explicitly, so you can see exactly which survive. When the shift is $k = 1$ only the first positive and the last negative remain; when the shift is larger (as with $\dfrac{1}{r} - \dfrac{1}{r+2}$) two positives and two negatives survive. A reliable way to keep track is to line the terms up in a column, writing the positive part of each term on the left and the negative part on the right, so that each negative sits directly above the positive it cancels. Anything that has no partner is a surviving term. With a shift of $2$, for instance, the $-\tfrac{1}{3}$ from the first term is cancelled by the $+\tfrac{1}{3}$ from the third term, not the second, so you must look two rows down; this is exactly why two terms survive at each end. Counting the survivors carefully is where most of the accuracy marks are won or lost, so it is worth the extra line of working even when the pattern looks obvious.

The sum to infinity

Once the sum is written in terms of $n$, let $n \to \infty$. The terms involving $n$ in the denominator tend to zero, so the sum to infinity is whatever start-terms remain. This connects the method to convergence and to improper integrals: a telescoping series converges precisely when the surviving end-terms approach a finite limit, which mirrors the way an improper integral converges when its limiting value is finite. In a typical exam answer you state the finite sum in terms of $n$, observe that the $n$-dependent fractions vanish as $n \to \infty$, and quote the limit as the sum to infinity, taking care to keep every constant start-term.

The method of differences relies on partial fractions and links to the sum-to-infinity idea, complementing the standard summation formulae.

Try this

Q1. Given $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$, state the sum to infinity of $\displaystyle\sum_{r=1}^{\infty}\dfrac{1}{r(r+1)}$. [2 marks]

• Cue. The finite sum is $1 - \tfrac{1}{n+1} \to 1$ as $n \to \infty$.

Q2. Write $\dfrac{1}{r} - \dfrac{1}{r+1}$ as a single fraction. [1 mark]

• Cue. $\dfrac{(r+1) - r}{r(r+1)} = \dfrac{1}{r(r+1)}$.