Telescoping Products

Products can telescope just as sums do, but through cancelling matching factors from the numerator and denominator instead of matching pieces of a difference. Recognizing a telescoping product, and sometimes factoring a single term to reveal two separate telescoping chains inside it, turns a long multiplication into a short one.

What you will learn

• Recognize a product that collapses through cancellation
• Rewrite a general factor to expose the cancellation
• Evaluate a finite telescoping product

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Recognizing a telescoping product

A product telescopes when each factor's numerator cancels with the denominator of a neighboring factor. In $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}$, the 2 in the first factor's denominator cancels with the 2 in the second factor's numerator, the 3 in the second factor's denominator cancels with the 3 in the third factor's numerator, and so on — leaving only the very first numerator and the very last denominator: $\frac{1}{6}$.

The same cancellation works in reverse for a product like $\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}$, where every middle value cancels between a denominator and the next numerator, leaving the last numerator over the first denominator: $\frac{6}{1}=6$.

Rewriting a factor to expose cancellation

Some factors do not look like a telescoping fraction until they are factored. The factor $1-\frac{1}{k^2}$ rewrites as $\frac{k^2-1}{k^2}$, and since $k^2-1$ is a difference of squares, this factors further as $\frac{(k-1)(k+1)}{k^2}=\frac{k-1}{k}\cdot\frac{k+1}{k}$ — splitting one factor into the product of two separate simple fractions.

Multiplying many such factors together separates into two independent telescoping chains: one chain built from all the $\frac{k-1}{k}$ pieces, which telescopes on its own, and another built from all the $\frac{k+1}{k}$ pieces, which telescopes separately. Evaluating each chain individually and then multiplying the two results together gives the value of the whole product.

Evaluating the full product carefully

As with telescoping sums, it helps to write out the first two and last two factors explicitly before trusting the general pattern, since the exact starting and ending index determines exactly which numerator and denominator survive. For $\prod_{k=a}^{b}\frac{k}{k+1}$, the surviving value is $\frac{a}{b+1}$ — the very first numerator over one more than the last denominator's base value.

When a product splits into two separate chains, as with the $1-\frac{1}{k^2}$ factors, each chain should be evaluated using this same first-over-last rule on its own index range, and the two results multiplied together only at the very end.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Multiplying every factor directly instead of noticing the numerators and denominators cancel across the product.
• Forgetting to split a factor like $1-\frac{1}{k^2}$ into two separate chains before attempting to telescope it.
• Mixing up which chain's first value and which chain's last value survive when a product splits into two telescoping pieces.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• A product telescopes when each factor's numerator cancels with a neighboring factor's denominator, leaving only the outermost values.
• A factor like $1-\frac{1}{k^2}$ factors via a difference of squares into $\frac{k-1}{k}\cdot\frac{k+1}{k}$, splitting into two separate telescoping chains.
• Track the surviving first and last values for each chain carefully, and multiply separate chains' results together only at the end.