Telescoping Sums and Products

An interviewer writes $\dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \dfrac{1}{3\cdot4} + \cdots + \dfrac{1}{99\cdot100}$ on the whiteboard and asks for the exact value. Adding ninety-nine fractions by hand is out of the question in an interview, and there's no obvious pattern to guess from the first couple of terms. The trick is to notice that each term isn't really a fraction to be added — it's a difference in disguise, and once you rewrite it that way almost everything in the sum cancels.

The idea: rewrite each term as a difference

A sum telescopes when you can write the general term $a_k$ as $b_k - b_{k+1}$ for some simpler sequence $b_k$. Then:

$$\sum_{k=1}^{n} (b_k - b_{k+1}) = (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_n - b_{n+1}) = b_1 - b_{n+1} .$$

In plain English: every middle term appears once with a plus sign and once with a minus sign from the neighboring pair, so it cancels — like a collapsible telescope sliding shut, only the two ends survive. A sum of $n$ terms becomes a subtraction of two numbers. The same idea works for products: if each factor is a ratio $b_k / b_{k+1}$, the product collapses to $b_1 / b_{n+1}$.

Worked example 1: the fraction sum

Notice that $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ — check it by combining the right side over a common denominator: $\dfrac{(k+1)-k}{k(k+1)} = \dfrac{1}{k(k+1)}$. So the sum becomes:

$$\sum_{k=1}^{99} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac{1}{99} - \frac{1}{100}\right) .$$

Every term from $\frac12$ through $\frac{1}{99}$ cancels against its neighbor, leaving only the very first piece and the very last piece: $1 - \dfrac{1}{100} = \dfrac{99}{100}$. Ninety-nine additions collapsed into one subtraction.

Worked example 2: a telescoping product

Evaluate $\displaystyle\prod_{k=2}^{10} \left(1 - \frac{1}{k^2}\right)$. Factor each term with a difference of squares: $1 - \dfrac{1}{k^2} = \dfrac{(k-1)(k+1)}{k^2} = \dfrac{k-1}{k}\cdot\dfrac{k+1}{k}$. Splitting the product into two separate telescoping chains, $\prod \frac{k-1}{k}$ runs $\frac12\cdot\frac23\cdots\frac{9}{10} = \frac{1}{10}$ (everything but the first numerator and last denominator cancels), and $\prod \frac{k+1}{k}$ runs $\frac32\cdot\frac43\cdots\frac{11}{10} = \frac{11}{2}$. Multiplying the two surviving pieces: $\frac{1}{10}\cdot\frac{11}{2} = \frac{11}{20}$.