Telescoping sums and inductive proof

Telescoping series are a powerful tool in the HSC Sequences and Series topic (Ext1). By decomposing each term into a difference using partial fractions, we can collapse a long sum into just a couple of boundary terms. In this problem, we'll walk through that technique and then lock in our result with mathematical induction, a core proof method in the syllabus.

Problem Statement

Consider the sequence

$$a_k=\frac{2}{k(k+1)(k+2)}\qquad (k\ge 1).$$

Define the partial sums

$$S_n=\sum_{k=1}^n a_k.$$

• Show, using partial fractions, that

$$\frac{2}{k(k+1)(k+2)} =\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}.$$

• Demonstrate that the sequence of partial sums telescopes. Hence find a simplified expression for $S_n$ in terms of $n$.
• Prove the expression for $S_n$ found in part (ii) by mathematical induction for all $n\ge 1$.
• Determine the limiting sum of the series as $n\to\infty$.

---

Hints

• For (i): Put the two fractions over the common denominator $k(k+1)(k+2)$.
• For (ii): Write out the first few terms of $S_n$ and look for cancellation.
• For (iii): Use $S_{m+1}=S_m+a_{m+1}$ in the induction step.
• For (iv): Take the limit of the closed form for $S_n$.

---

Solutions

(i) To verify the identity, we start with the right-hand side and combine the fractions over the common denominator $k(k+1)(k+2)$.

$$\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)} =\frac{k+2-k}{k(k+1)(k+2)} =\frac{2}{k(k+1)(k+2)}.$$

(ii) Using the decomposition from part (i), we can rewrite each term $a_k$ as a difference. Then writing out the sum $S_n$ reveals that all interior terms cancel, leaving only the first positive part and the last negative part. So

$$a_k=\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}.$$

So

$$\begin{aligned} S_n &=\left(\frac1{1\cdot2}-\frac1{2\cdot3}\right) +\left(\frac1{2\cdot3}-\frac1{3\cdot4}\right) +\cdots +\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right)\\ &=\frac12-\frac{1}{(n+1)(n+2)}. \end{aligned}$$

(iii) Mathematical induction is the standard method to confirm the closed form we found. We check the base case, assume the statement for $n=m$, and then use $S_{m+1}=S_m+a_{m+1}$ together with the telescoping observation to prove it for $m+1$. We prove

$$S_n=\frac12-\frac{1}{(n+1)(n+2)}.$$

For $n=1$,

$$S_1=\frac{2}{1\cdot2\cdot3}=\frac13$$

and

$$\frac12-\frac1{2\cdot3}=\frac12-\frac16=\frac13.$$

Now assume the formula holds for $n=m$. Then

$$\begin{aligned} S_{m+1} &=S_m+\frac{2}{(m+1)(m+2)(m+3)}\\ &=\frac12-\frac1{(m+1)(m+2)} +\left(\frac1{(m+1)(m+2)}-\frac1{(m+2)(m+3)}\right)\\ &=\frac12-\frac1{(m+2)(m+3)}. \end{aligned}$$

This is the required formula with $n=m+1$, so the result holds for all $n\ge1$.

(iv) Finally, with the finite sum formula at hand, taking the limit as $n\to\infty$ is straightforward—the fraction with $n$ in the denominator vanishes.

$$\lim_{n\to\infty}S_n =\lim_{n\to\infty}\left(\frac12-\frac{1}{(n+1)(n+2)}\right) =\frac12.$$

---

Takeaways

• Telescoping sums work by rewriting terms so that most neighbouring parts cancel.
• Induction is a useful way to verify a closed form found by pattern spotting.
• A finite partial-sum formula can make the limiting sum immediate.

---

Further Readings

HSC Collections, HSC Vectors, HSC Polynomials, HSC Functions