Overview

Many competition problems ask you to evaluate a sum or product whose closed form is not immediately obvious. A powerful strategy is telescoping: rewriting each term so that most of it cancels with a neighboring term, leaving only a handful of boundary terms.

The name comes from a collapsible spyglass — you start with something long and it collapses down to something short. The workflow is always:

1. Decompose the general term into a difference (for sums) or ratio (for products).
2. Write out several terms to confirm the cancellation pattern.
3. Collapse down to boundary terms.
4. Evaluate what remains.

Telescoping Sums

The Fundamental Idea

If you compute $\sum_{k=1}^{n}(a_k - a_{k+1})$, writing it out gives

Every intermediate term cancels. The challenge is finding the decomposition that turns your expression into this form.

Partial Fractions

The most common tool is partial fraction decomposition. For $\frac{1}{k(k+1)}$, we solve $\frac{1}{k(k+1)} = \frac{A}{k}+\frac{B}{k+1}$, giving $A=1$, $B=-1$:

So $\sum_{k=1}^{n}\frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

Wider Gaps

When factors are not consecutive, cancellation still occurs but with a gap. For $\frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$, writing out terms:

Each positive $\frac{1}{m}$ cancels with the negative $\frac{1}{m}$ appearing two terms later. The first 2 and last 2 terms survive:

General principle: a decomposition with gap $d$ leaves $d$ boundary terms at each end.

Quadratic Denominators

Since $k^2-1=(k-1)(k+1)$, we get gap 2:

So $\sum_{k=2}^{n}\frac{1}{k^2-1} = \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right)$, which converges to $\frac{3}{4}$ as $n\to\infty$.

Rationalizing Radicals

For $\frac{1}{\sqrt{k}+\sqrt{k+1}}$, multiply by the conjugate:

Then $\sum_{k=1}^{n}\frac{1}{\sqrt{k}+\sqrt{k+1}} = \sqrt{n+1}-1$.

Double Telescoping (Cubic Denominators)

For $\frac{1}{k(k+1)(k+2)}$, partial fractions give

Defining $f(k)=\frac{1}{k}-\frac{1}{k+1}$, each term is $\frac{1}{2}(f(k)-f(k+1))$, which telescopes to

Telescoping Products

The multiplicative analogue: $\prod_{k=1}^{n}\frac{a_{k+1}}{a_k} = \frac{a_{n+1}}{a_1}$, since each numerator cancels with the adjacent denominator.

Example. $\prod_{k=2}^{n}\frac{k-1}{k} = \frac{1}{2}\cdot\frac{2}{3}\cdots\frac{n-1}{n} = \frac{1}{n}$.

Example. For $\prod_{k=2}^{n}\frac{k^2-1}{k^2}$, factor as $\frac{(k-1)(k+1)}{k\cdot k}$ and split into two products:

Strategy Checklist

• Factor the denominator and look for consecutive integer factors.
• Apply partial fractions to express each term as a difference.
• Write out 4–5 terms explicitly — do not skip this.
• Identify the gap $d$ and track $d$ boundary terms at each end.
• For products, rewrite each factor as $\frac{a_{k+1}}{a_k}$.
• For infinite sums, verify the tail terms vanish.

Common Pitfalls

• Forgetting boundary terms. With gap $d$, you must keep $d$ terms at each end. Losing one is the most common error.
• Expanding before factoring. Multiplying out $(k-1)(k+1)=k^2-1$ just to re-factor it wastes effort and invites mistakes.
• Mismatched indices. Verify that $\frac{1}{k+1}$ from term $k$ actually cancels with $\frac{1}{k+1}$ from the correct neighboring term.
• Ignoring convergence. For infinite series, confirm $a_n\to 0$.

Practice Problems

Problem: 2025 AMC 10B Problem 15

The sum $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^3+6k^2+8k}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. What is $a+b$?

$\textbf{(A)}\ 89\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 107\qquad\textbf{(E)}\ 129$

Solution

Factor the denominator. $k^3+6k^2+8k = k(k+2)(k+4)$.

Partial fractions. Setting $\frac{1}{k(k+2)(k+4)}=\frac{A}{k}+\frac{B}{k+2}+\frac{C}{k+4}$ and solving (plug in $k=0,-2,-4$) gives $A=\frac{1}{8}$, $B=-\frac{1}{4}$, $C=\frac{1}{8}$:

$$\frac{1}{k(k+2)(k+4)} = \frac{1}{8}\left(\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}\right).$$

Regroup for telescoping. Write $\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}$ as

$$\left(\frac{1}{k}-\frac{1}{k+2}\right)-\left(\frac{1}{k+2}-\frac{1}{k+4}\right) = f(k)-f(k+2),$$

where $f(k)=\frac{1}{k}-\frac{1}{k+2}$. This is a gap-2 telescope in $f$.

Collapse. $\sum_{k=1}^{\infty}(f(k)-f(k+2)) = f(1)+f(2)$ since $f(k)\to 0$.

$$f(1)+f(2) = \left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right) = \frac{2}{3}+\frac{1}{4} = \frac{11}{12}.$$

Final answer. $\frac{1}{8}\cdot\frac{11}{12} = \frac{11}{96}$, and $\gcd(11,96)=1$, so $a+b = 11+96 = \boxed{107}$. The answer is $\textbf{(D)}$.