Telescoping sums: how to spot them before you compute

A telescoping sum is the cheapest trick in competition algebra and the one most often missed. Not because the idea is hard, but because nothing on the page says "telescope". You are handed a sum with ninety-nine terms and a clock, and the only question that matters is whether the middle of it is going to die.

So this is not an explanation of what telescoping is. It is a list of the things on the page that should make you stop and look for the collapse before you try anything else.

The four signals

• A denominator that is a product of two factors a fixed distance apart. $\frac{1}{k(k+1)}$, $\frac{1}{k(k+2)}$, $\frac{1}{(2k-1)(2k+1)}$. This is the single most common tell in the whole subject. Partial fractions turns any of them into a difference, and a difference is a telescope.
• A sum with an absurd number of terms and an exact answer expected. Nobody asks you to add ninety-nine fractions by hand. If the problem names a specific large upper limit and still wants a closed form, the terms in the middle are decoration and the setter knows it.
• A term you can already read as a difference. Anything of the shape $a_{k+1} - a_k$ is a telescope wearing no disguise, and recurrences hand you these constantly. Fibonacci is the standard case: the recurrence rearranges to $F_k = F_{k+2} - F_{k+1}$, so summing the sequence collapses to $F_{n+2} - 1$.
• A product of ratios. The same collapse happens multiplicatively when each factor looks like $\frac{f(k+1)}{f(k)}$. The classic is $\prod_{k=2}^{n}\left(1 - \frac{1}{k^2}\right)$, which factors into $\frac{(k-1)(k+1)}{k^2}$ and slides down to $\frac{n+1}{2n}$.

There is a fifth signal worth its own line because it is disguised: a sum of reciprocals of surds. $\frac{1}{\sqrt{k}+\sqrt{k+1}}$ looks like nothing until you rationalise, at which point it is exactly $\sqrt{k+1}-\sqrt{k}$ and the sum is $\sqrt{n+1}-1$. Rationalising a denominator is normally cosmetic. Here it is the whole solution.

The three questions that turn a signal into an answer

Spotting it is most of the work. Finishing it is where people lose the marks, and they lose them in the same three places every time.

1. What is the function? Write down the $f$ for which your term equals $f(k) - f(k+1)$, explicitly, on its own line. If you cannot name $f$, you have a hunch, not a telescope.
2. What is the step? The factors in $\frac{1}{k(k+2)}$ are two apart, not one, so the cancellation skips a term. Two survive at the front and two at the back, and partial fractions leaves a $\frac{1}{2}$ outside that people drop.
3. What actually survives? Write out the first two terms and the last two terms in full, with real numbers in them. Never reason about the ends from the ellipsis. Almost every wrong telescoping answer is off by one boundary term, and the ellipsis is where that error hides.

Question three is a habit, not an insight, and it is the one worth drilling. Two minutes of writing $\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac{1}{99}-\frac{1}{100}\right)$ out properly beats twenty minutes of staring at a general term.

When it looks like a telescope and is not

The famous near-miss is $\sum \frac{1}{k^2}$. It has the rigid shape, it has the reciprocal, and it does not telescope: there is no nice $f$ with $\frac{1}{k^2} = f(k) - f(k+1)$.

What you do instead is the move worth stealing from this whole topic. Bound each term by one that does telescope. Since $\frac{1}{k^2} < \frac{1}{(k-1)k} = \frac{1}{k-1} - \frac{1}{k}$ for $k \ge 2$, the sum is under $2 - \frac{1}{n}$, and you have a proof of boundedness in one line without ever finding the exact value.

That is the real lesson. Telescoping is not only a way to evaluate a sum; it is a way to bound one, and the bounding version shows up in far more olympiad problems than the exact version does. If a sum resists collapsing, do not abandon the technique. Ask what nearby sum does collapse and whether an inequality gets you there.

The other thing to keep in mind is the relationship with induction. Once you have telescoped a sum you have a closed form, and a closed form can always be verified by induction afterwards. The two are not rivals. Telescoping is how you find the answer; induction is how you defend it when a proof is demanded rather than a number.

What to practise

Recognition is trained by sorting, not by solving. Take fifteen sums, and for each one write a single word: telescope, or not. Do not evaluate any of them. You will get through fifteen in the time one full computation takes, and the sorting is the skill you are actually short of on the day.

Lemma teaches this as a named technique rather than as a trick you meet once and forget. The telescoping page has the statement, the product version, and where it sits in the levels. The problem archive holds the proof-shaped versions with difficulty and expected time on each, and the daily problem posts a new one every day in three tiers, with no account needed.