Surd telescoping

In this post, we explore the surd telescoping technique — a clever application of conjugate rationalisation and series cancellation that appears in the HSC sequences syllabus. By mastering this method, you'll gain insight into simplifying surd sums, finding closed forms, and rapidly solving inequality problems.

Problem Statement

Let

$$a_k=\frac1{\sqrt{k+1}+\sqrt{k}},\qquad S_n=\sum_{k=1}^n a_k.$$

• Show $a_k=\sqrt{k+1}-\sqrt{k}$.
• Find $S_n$.
• Evaluate $S_{99}$.
• Find the smallest integer $N$ such that $S_N>10$.

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Hints

Rationalize with the conjugate and then telescope.

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Solutions

We begin by rationalising the denominator using the conjugate, a standard trick to eliminate surds from the denominator:

$$a_k=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}.$$

Because each $a_k$ can be expressed as a difference of consecutive square roots, the sum telescopes beautifully: all intermediate terms cancel, leaving only the first negative and last positive terms.

Hence

$$S_n=(\sqrt2-\sqrt1)+\cdots+(\sqrt{n+1}-\sqrt n)=\sqrt{n+1}-1.$$

So $S_{99}=10-1=9$. For $S_N>10$:

We set up an inequality using the closed form and solve it algebraically.

$$\sqrt{N+1}-1>10 \implies N>120,$$

smallest such integer is $N=121$.

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Takeaways

• Rationalising the denominator with a conjugate turns a fraction of surds into a simpler difference, enabling telescoping.
• Telescoping sums collapse to just the end terms, drastically simplifying evaluation.
• Once you have a closed form, solving equations or inequalities becomes a straightforward algebraic task.

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Further Readings

HSC Functions, HSC Complex Numbers, HSC Vectors, HSC Polynomials