Alternative telescoping patterns

Telescoping series are one of the most elegant patterns in the HSC Sequences and Series topic—they reveal how careful algebraic manipulation can collapse a long sum into just two terms. The problem that follows begins with a concrete alternating sum, then builds the classic partial fraction decomposition that unlocks telescoping. By working through all three parts, you’ll see exactly how the cancellation mechanism works and gain a potent tool for both finite sums and infinite limits.

Problem Statement

• Evaluate $\sum_{n=0}^{4}(-1)^{n+1}(n+5)$.
• Show $\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}$.
• Hence find

$$S_N=\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)$$

and $\lim_{N\to\infty}S_N$.

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Hints

Part (iii) is a direct telescoping sum after part (ii).

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Solutions

For part (i), because the sum only has five terms, direct expansion is the quickest approach.

$$\sum_{n=0}^{4}(-1)^{n+1}(n+5)=-7.$$

The trick for part (ii) is to notice that the numerator can be expressed as the difference of the two factors in the denominator, which immediately splits the fraction.

$$\frac1{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac1n-\frac1{n+1}.$$

With this decomposition, $S_N$ becomes a classic telescoping series. Expanding the brackets shows that all intermediate terms cancel, leaving only the very first and very last.

$$S_N=1-\frac1{N+1},\qquad \lim_{N\to\infty}S_N=1.$$

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Takeaways

• Finite alternating sums can be evaluated by direct expansion—always check if the number of terms is small enough to do this safely.
• The identity $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ is a fundamental partial fraction that converts a rational expression into a telescoping difference.
• When a series is telescoping, $S_N$ simplifies to $a_1 - a_{N+1}$; the infinite limit then follows immediately by letting the last term tend to zero.

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Further Readings

HSC Combinatorics, HSC Proofs, HSC Polynomials, HSC Inequalities