Telescoping is the habit of arranging sums or products so most terms cancel and only boundary terms remain. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
When each term of a sum can be written as a difference of consecutive values of one function, the interior terms cancel in pairs and only the endpoints survive.
Why It Works
1
Write the sum out: $\big(f(2) - f(1)\big) + \big(f(3) - f(2)\big) + \cdots + \big(f(n+1) - f(n)\big)$.
2
Every $+f(k)$ from one term is cancelled by the $-f(k)$ in the next term.
3
All that remains is the last positive term $f(n+1)$ and the first negative term $-f(1)$.
4
So the sum collapses to $f(n+1) - f(1)$. The trick is recognizing or engineering the difference form $f(k+1) - f(k)$ for each summand.
Split into two telescoping products: $\prod_{k=2}^{n}\dfrac{k-1}{k}\cdot \prod_{k=2}^{n}\dfrac{k+1}{k}$. The first is $\dfrac{1}{2}\cdot\dfrac{2}{3}\cdots\dfrac{n-1}{n} = \dfrac{1}{n}$; the second is $\dfrac{3}{2}\cdot\dfrac{4}{3}\cdots\dfrac{n+1}{n} = \dfrac{n+1}{2}$.
3
Multiply: $\dfrac{1}{n}\cdot\dfrac{n+1}{2} = \dfrac{n+1}{2n}$. At $n = 10$: $\dfrac{11}{20}$.
Answer:
$\displaystyle\prod_{k=2}^{n}\left(1 - \frac{1}{k^2}\right) = \frac{n+1}{2n}$; at $n = 10$ it is $\dfrac{11}{20}$.
Going Deeper
Generalization: telescoping is the discrete Fundamental Theorem of Calculus — summing a difference $\Delta f(k) = f(k+1) - f(k)$ recovers $f$ at the endpoints. The same idea telescopes PRODUCTS (sum of logs) and works for any window width by writing $f(k + m) - f(k)$.
Where it appears: AIME finite-sum evaluations, partial-fraction sums like $\sum \frac{1}{k(k+1)}$, factorial identities, and any product of ratios that chains; spotting or engineering the difference form is the whole game.
Pitfall: track the surviving boundary terms EXACTLY — with a width-$m$ telescope, $m$ terms remain at each end, not one. Off-by-one errors on the first/last index (e.g. starting at $k = 2$ instead of $k = 1$) are the dominant mistake; always write out the first two and last two terms explicitly to confirm what cancels.
Spot the Signal
Look for problems where the key step is arranging sums or products so most terms cancel and only boundary terms remain.
You can describe the hard part as arranging sums or products so most terms cancel and only boundary terms remain, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the expression or equation that calls for telescoping, then rewrite the givens around it.
Name the relation that makes Telescoping legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Telescoping just because the surface looks familiar; verify the required condition first.
Applying Telescoping because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let telescoping reveal the algebraic structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is arranging sums or products so most terms cancel and only boundary terms remain.