I’m stuck on a telescoping series and my brain keeps doing the math equivalent of tripping over its own shoelaces.
I’m looking at S = Σ (from n=1 to ∞) of 1/(n(n+1)). I broke it up as 1/n − 1/(n+1), which feels like the classic “domino effect” setup. So I wrote out the first few terms:
(1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + …
Everything cancels in pairs, right? Like matching every expense with a refund. So I concluded S = 0 because each +1/k is canceled by the next −1/k, and at the end there’s nothing left. I even told myself that the last leftover bit is −1/(∞) which is 0, so the initial 1 also disappears. That sounds super neat to me-like eating a pizza where every slice is instantly replaced by a negative slice until the plate’s empty.
But something about this feels too good to be true. Where is my cancellation logic breaking down? Is it actually valid to say everything cancels in an infinite sum like this, and to treat 1/(∞) as 0 to wipe out the first term? Or am I pairing things in a way that’s not allowed?
Any help appreciated!
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3 Responses
Great instinct to split 1/(n(n+1)) into 1/n − 1/(n+1)! That’s exactly the telescoping move. The safe way to use it is with partial sums: let S_N = Σ_{n=1}^N (1/n − 1/(n+1)). Everything in the middle cancels, leaving S_N = 1 − 1/(N+1). Now take the limit as N → ∞: S = lim S_N = 1 − 0 = 1. So the series sums to 1, not 0.
Where did the “everything cancels to 0” idea slip? In infinite sums you can’t cancel “all at once”; you must cancel inside a finite partial sum and then take a limit. In S_N every internal +1/k meets a −1/k, but there is no −1/1 to cancel the very first +1/1. The leftover is 1 − 1/(N+1), and 1/(N+1) tends to 0 (it’s not literally 1/∞). Bonus sanity check: each term 1/(n(n+1)) is positive, so the partial sums are increasing and can’t possibly head down to 0.
Analogy time: think of a collapsible spyglass. You slide all the inner tubes in and almost everything vanishes-but one end cap stays in your hand (that’s the “1”), while the other end cap retreats farther and farther away until it’s effectively out of sight (that’s the −1/(N+1) term shrinking to 0). The middle tubes cancel perfectly, but that first cap survives, so the total settles at 1.
You’re so close! The split 1/(n(n+1)) = 1/n − 1/(n+1) is exactly right-the only place the logic slips is treating the infinite cancellation as if there were a “last term.” The safe way is to look at the Nth partial sum: S_N = (1 − 1/2) + (1/2 − 1/3) + … + (1/N − 1/(N+1)) = 1 − 1/(N+1). Now let N go to infinity: 1/(N+1) → 0, so the full sum is 1, not 0. The initial 1 never gets canceled because there’s no earlier −1/1 waiting to pounce; in a telescope you always have a left edge that survives. Saying “−1/(∞)” is 0 is a tempting story, but it isn’t an actual term-only the limit of the right-end leftover. So everything cancels except the first 1, and the tiny tail −1/(N+1) fades to 0 in the limit. I might be overexplaining, but that’s the whole trick: do the cancellation on finite partial sums first, then take the limit, and you get S = 1.
Your decomposition is right: 1/(n(n+1)) = 1/n − 1/(n+1). The safe way with an infinite series is to look at partial sums: S_N = ∑_{n=1}^N (1/n − 1/(n+1)) = (1 − 1/2) + (1/2 − 1/3) + … + (1/N − 1/(N+1)) = 1 − 1/(N+1). Then take the limit: as N → ∞, 1/(N+1) → 0, so S = lim S_N = 1. The cancellation intuition breaks only if you imagine a “last” negative term that cancels the very first 1; there isn’t one. The symbol 1/∞ isn’t an actual term in the sum-only the limit of 1/(N+1), and that limit affects the tail, not the initial 1. Think of it like lining up charges and coupons where each coupon cancels the next charge, but there is no final coupon to wipe out the very first charge; that first one survives, giving the total 1.