Telescoping Series Calculator: Exact Sums by Mass Cancellation
About the Author
Marko Šinko
Co-Founder & Lead Developer, AI Math Calculator
Ask a telescoping series calculator for Σ 1/(n(n+1)) and it hands back exactly 1. Not 0.9999, not “converges by comparison” — the number 1, as a clean fraction. That’s the rare privilege of this family: while most convergent series only surrender approximations, a telescoping sum collapses algebraically, leaving two or three fractions you can add by hand. The catch is that the collapse has rules. Which terms survive depends on the gap between the denominator factors, a detail that produces more wrong homework answers than the calculus itself. The tool above does the exact arithmetic and shows the cancellation term by term; this page explains the method, tabulates the classic sums, and covers the two situations where a telescoping series refuses to behave.
Split, Cancel, Take the Limit
The quick answer, in three moves. First, split the general term into a difference of the form f(n) − f(n+d). For rational terms that’s a partial fraction decomposition: 1/(n(n+1)) = 1/n − 1/(n+1). Second, write out the partial sum S(N) and let the interior terms annihilate each other — the +1/2 from one bracket meets the −1/2 from the previous bracket, and so on down the line. Third, take N → ∞ in whatever survives. For 1/(n(n+1)) the survivors are 1/1 at the front and −1/(N+1) at the back, so S(N) = 1 − 1/(N+1) → 1.
The name comes from those collapsible brass spyglass telescopes: a long chain of segments slides together into something short. What makes the trick powerful is the third step. Because S(N) has a closed form, you get the sum itself — not just a convergence verdict, which is all a limit comparison test would ever tell you.
Watching 1/(n(n+1)) Collapse, Term by Term
Here is the whole mechanism for the classic example, written out to five terms:
S(5) = (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + (1/4 − 1/5) + (1/5 − 1/6)
= 1/1 − 1/6 = 5/6
Every negative fraction is destroyed by the positive copy of itself in the next bracket. Only the very first positive term and the very last negative term have no partner. That gives the general formula S(N) = 1 − 1/(N+1) with no summation left in it — try N = 5 and you get 5/6, matching the expansion above. S(100) = 100/101, S(1000) = 1000/1001. The partial sums crawl up toward 1 and the leftover gap is exactly 1/(N+1), which is why the table in the calculator shrinks its “gap” column by a factor of 10 each row. If you only need a numeric partial sum for a series that doesn’t telescope, a general summation calculator is the better tool — the trick here only pays off when the algebra cancels.
The Gap Rule: What Survives When b − a > 1
Now the detail that separates an A from a B on this topic. Take Σ 1/(n(n+2)). Partial fractions give (1/2)(1/n − 1/(n+2)) — the difference now jumps by 2, so each negative term must wait two brackets for its cancelling partner. The consequence: two leading terms survive instead of one. The sum is (1/2)(1/1 + 1/2) = 3/4, not the (1/2)(1/1) = 1/2 that students write when they assume only the first term lives.
The general rule: if the term splits as k·[f(n) − f(n+d)], then exactly d leading values of f survive, and d trailing values appear in every partial sum. Starting at n = n₀,
Σ = k · [f(n₀) + f(n₀+1) + … + f(n₀+d−1)] (when f(n) → 0)
For 4/((n+3)(n+7)) the gap is 4, so four fractions survive. The constant out front is 1/(b−a) = 1/4, which the numerator 4 cancels exactly, leaving 1/4 + 1/5 + 1/6 + 1/7 = 319/420 ≈ 0.7595. Punch it into the calculator and watch the cascade: the first four positive terms glow green with no partner in sight.
Six Telescoping Sums Worth Knowing Cold
These six cover nearly every telescoping problem that appears on Calc 2 exams. All sums start at n = 1.
| Series | Telescoping form | Exact sum |
|---|---|---|
| Σ 1/(n(n+1)) | 1/n − 1/(n+1) | 1 |
| Σ 1/(n(n+2)) | ½[1/n − 1/(n+2)] | 3/4 |
| Σ 1/((2n−1)(2n+1)) | ½[1/(2n−1) − 1/(2n+1)] | 1/2 |
| Σ 1/(n(n+1)(n+2)) | ½[1/(n(n+1)) − 1/((n+1)(n+2))] | 1/4 |
| Σ arctan(1/(n²+n+1)) | arctan(n+1) − arctan(n) | π/4 |
| Σ ln((n+1)/n) | ln(n+1) − ln(n) | diverges |
Notice the third row: the factors are 2n−1 and 2n+1, not n+a and n+b, yet the mechanism is identical — the substitution m = 2n−1 turns it into consecutive odd denominators. And the fourth row telescopes on pairs: take f(n) = 1/(2n(n+1)) and the triple-product term is exactly f(n) − f(n+1).
Yes, a Telescoping Series Can Diverge
The last row of the table deserves its own warning, because “it telescopes, so it converges” is a surprisingly common false belief. The partial sum of Σ ln((n+1)/n) collapses beautifully: S(N) = ln(N+1) − ln(1) = ln(N+1). The collapse worked perfectly — and the result marches off to infinity anyway, just slowly. Same story for Σ (√(n+1) − √n), whose partial sum is √(N+1) − 1.
The honest statement: a telescoping series Σ [f(n) − f(n+1)] converges exactly when lim f(n) exists and is finite, and then the sum equals f(n₀) − lim f(n). For 1/n that limit is 0 and everything works. For ln(n) and √n the limit is infinite, so the series diverges even though every interior term cancelled. When the limit of f is hard to see, run the series through a series convergence calculator first — knowing the verdict before you hunt for the sum saves real time. Rational terms like c/((n+a)(n+b)) are always safe: they behave like c/n², a convergent p-series with p = 2, which is why the calculator on this page always reports convergence.
Spotting the Telescope When the Problem Hides It
Exam writers rarely serve the difference form on a plate. Three disguises account for most problems. The rational disguise is a single fraction with a factored (or factorable) quadratic underneath: 1/(n² + 3n + 2) looks unremarkable until you factor it as 1/((n+1)(n+2)) and split. Anything of the form c/(quadratic in n) with real integer roots should trigger the partial-fractions reflex.
The logarithm disguise hides the difference inside one log: Σ ln(n(n+2)/(n+1)²) doesn’t look telescoping until you split it as [ln(n) − ln(n+1)] − [ln(n+1) − ln(n+2)] — a telescope of telescopes that converges to −ln 2 ≈ −0.693. The arctan disguise leans on the identity arctan(x) − arctan(y) = arctan((x−y)/(1+xy)): the innocent-looking Σ arctan(1/(n²+n+1)) is really Σ [arctan(n+1) − arctan(n)], which collapses to π/2 − π/4 = π/4 ≈ 0.785. The tell in every disguise is the same — a parameter appearing twice, shifted. If substituting n → n+1 into part of the term reproduces another part, you’re holding a telescope.
Off-by-One Errors a Telescoping Series Calculator Catches
Grading pattern from years of these problems: the calculus is almost never the mistake — the bookkeeping is. Four errors dominate. Assuming one survivor when the gap is 2 turns 3/4 into 1/2 (the gap rule above). Dropping the trailing term from a finite sum inflates the answer: Σ from 1 to 20 of 1/(n(n+1)) is 20/21, not 1, because −1/21 is still alive at N = 20. Forgetting a shifted start costs the first term: Σ from n = 3 of 1/(n(n+1)) is 1/3, not 1, since the survivors are counted from n₀ = 3. And sign slips in the decomposition — writing 1/(n(n+1)) = 1/n + 1/(n+1) — produce a divergent harmonic-style sum that should fail your own sanity check within seconds, since the original terms behave like 1/n².
Each of these is visible instantly in the cascade view above: if your hand-derived survivors don’t match the green terms, one of the four errors is the reason. For sequences of partial sums and their limits more generally, the sequence calculator handles the S(N) side of the story.
