Intuition The one-line ideaA telescoping series is one where each term can be written as a difference of two pieces of a single sequence, so that when you add the terms, interior pieces cancel and only a few survive — just like a folding telescope collapsing into itself.
Most series are impossible to sum exactly — we only get convergence tests . Telescoping is one of the rare cases where we get the EXACT sum , not just "it converges." The trick: turn a sum into a problem of cancellation.
Definition Telescoping seriesA series ∑ n = 1 ∞ a n \sum_{n=1}^\infty a_n ∑ n = 1 ∞ a n telescoping if there exists a sequence b n b_n b n a n = b n − b n + 1 . a_n = b_n - b_{n+1}. a n = b n − b n + 1 . S N = ∑ n = 1 N a n = b 1 − b N + 1 . S_N = \sum_{n=1}^N a_n = b_1 - b_{N+1}. S N = ∑ n = 1 N a n = b 1 − b N + 1 . ⟺ lim N → ∞ b N + 1 \iff \lim_{N\to\infty} b_{N+1} ⟺ lim N → ∞ b N + 1 ∑ a n = b 1 − lim N → ∞ b N + 1 \sum a_n = b_1 - \lim_{N\to\infty}b_{N+1} ∑ a n = b 1 − lim N → ∞ b N + 1
WHAT we want: S N = ∑ n = 1 N ( b n − b n + 1 ) S_N = \sum_{n=1}^N (b_n - b_{n+1}) S N = ∑ n = 1 N ( b n − b n + 1 )
Write out every term — this is the whole game:
S N = ( b 1 − b 2 ) + ( b 2 − b 3 ) + ( b 3 − b 4 ) + ⋯ + ( b N − b N + 1 ) S_N = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \dots + (b_N - b_{N+1}) S N = ( b 1 − b 2 ) + ( b 2 − b 3 ) + ( b 3 − b 4 ) + ⋯ + ( b N − b N + 1 ) Why this step? I expand so I can physically see what touches what. Each − b k -b_{k} − b k k − 1 k-1 k − 1 + b k +b_k + b k k k k
Group the survivors and the cancellers:
S N = b 1 + ( − b 2 + b 2 ) ⏟ 0 + ( − b 3 + b 3 ) ⏟ 0 + ⋯ + ( − b N + b N ) ⏟ 0 − b N + 1 S_N = b_1 + \underbrace{(-b_2 + b_2)}_{0} + \underbrace{(-b_3 + b_3)}_{0} + \dots + \underbrace{(-b_N + b_N)}_{0} - b_{N+1} S N = b 1 + 0 ( − b 2 + b 2 ) + 0 ( − b 3 + b 3 ) + ⋯ + 0 ( − b N + b N ) − b N + 1 Why this step? Every interior b k b_k b k 2 ≤ k ≤ N 2\le k\le N 2 ≤ k ≤ N + + + − - − first b 1 b_1 b 1 last − b N + 1 -b_{N+1} − b N + 1
S N = b 1 − b N + 1 \boxed{S_N = b_1 - b_{N+1}} S N = b 1 − b N + 1 Take the limit: if b N + 1 → L b_{N+1}\to L b N + 1 → L ∑ n = 1 ∞ a n = b 1 − L \displaystyle\sum_{n=1}^\infty a_n = b_1 - L n = 1 ∑ ∞ a n = b 1 − L
Series rarely arrive pre-written as b n − b n + 1 b_n - b_{n+1} b n − b n + 1 manufacture it, usually via partial fractions.
∑ n = 1 ∞ 1 n ( n + 1 ) \displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)} n = 1 ∑ ∞ n ( n + 1 ) 1 Step 1 — split into a difference. Write 1 n ( n + 1 ) = A n + B n + 1 \dfrac{1}{n(n+1)} = \dfrac{A}{n} + \dfrac{B}{n+1} n ( n + 1 ) 1 = n A + n + 1 B Why? I want the form b n − b n + 1 b_n - b_{n+1} b n − b n + 1
Clearing denominators: 1 = A ( n + 1 ) + B n 1 = A(n+1) + Bn 1 = A ( n + 1 ) + B n n = 0 ⇒ A = 1 n=0\Rightarrow A=1 n = 0 ⇒ A = 1 n = − 1 ⇒ B = − 1 n=-1 \Rightarrow B=-1 n = − 1 ⇒ B = − 1 1 n ( n + 1 ) = 1 n − 1 n + 1 \frac{1}{n(n+1)} = \frac1n - \frac{1}{n+1} n ( n + 1 ) 1 = n 1 − n + 1 1 Step 2 — identify b n b_n b n Here b n = 1 n b_n = \dfrac1n b n = n 1 a n = b n − b n + 1 a_n = b_n - b_{n+1} a n = b n − b n + 1 Why? Confirms it telescopes — the − 1 n + 1 -\frac{1}{n+1} − n + 1 1 n n n + 1 n + 1 +\frac{1}{n+1} + n + 1 1 n + 1 n+1 n + 1
Step 3 — partial sum. S N = b 1 − b N + 1 = 1 − 1 N + 1 S_N = b_1 - b_{N+1} = 1 - \dfrac{1}{N+1} S N = b 1 − b N + 1 = 1 − N + 1 1
Step 4 — limit. lim N → ∞ ( 1 − 1 N + 1 ) = 1. \displaystyle\lim_{N\to\infty}\Big(1 - \tfrac{1}{N+1}\Big) = 1. N → ∞ lim ( 1 − N + 1 1 ) = 1. ∑ n = 1 ∞ 1 n ( n + 1 ) = 1 \boxed{\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1} n = 1 ∑ ∞ n ( n + 1 ) 1 = 1
Worked example Two-step gap:∑ n = 1 ∞ 1 n ( n + 2 ) \displaystyle\sum_{n=1}^\infty \frac{1}{n(n+2)} n = 1 ∑ ∞ n ( n + 2 ) 1 Step 1. 1 n ( n + 2 ) = 1 / 2 n − 1 / 2 n + 2 = 1 2 ( 1 n − 1 n + 2 ) \dfrac{1}{n(n+2)} = \dfrac{1/2}{n} - \dfrac{1/2}{n+2} = \tfrac12\left(\dfrac1n - \dfrac{1}{n+2}\right) n ( n + 2 ) 1 = n 1/2 − n + 2 1/2 = 2 1 ( n 1 − n + 2 1 ) Why? A = 1 2 , B = − 1 2 A=\frac12, B=-\frac12 A = 2 1 , B = − 2 1 1 = A ( n + 2 ) + B n 1=A(n+2)+Bn 1 = A ( n + 2 ) + B n
Step 2 — beware the gap of 2. This is NOT b n − b n + 1 b_n-b_{n+1} b n − b n + 1 b n − b n + 2 b_n-b_{n+2} b n − b n + 2 two terms survive at each end.
Write out:
S N = 1 2 [ ( 1 − 1 3 ) + ( 1 2 − 1 4 ) + ( 1 3 − 1 5 ) + ⋯ + ( 1 N − 1 N + 2 ) ] S_N=\tfrac12\Big[(1-\tfrac13)+(\tfrac12-\tfrac14)+(\tfrac13-\tfrac15)+\dots+(\tfrac1N-\tfrac1{N+2})\Big] S N = 2 1 [ ( 1 − 3 1 ) + ( 2 1 − 4 1 ) + ( 3 1 − 5 1 ) + ⋯ + ( N 1 − N + 2 1 ) ] + 1 3 +\frac13 + 3 1 − 1 3 -\frac13 − 3 1 start : 1 1 1 1 2 \frac12 2 1 end : − 1 N + 1 , − 1 N + 2 -\frac{1}{N+1}, -\frac{1}{N+2} − N + 1 1 , − N + 2 1 S N = 1 2 ( 1 + 1 2 − 1 N + 1 − 1 N + 2 ) S_N = \tfrac12\Big(1 + \tfrac12 - \tfrac{1}{N+1} - \tfrac{1}{N+2}\Big) S N = 2 1 ( 1 + 2 1 − N + 1 1 − N + 2 1 ) Step 3 — limit. ∑ = 1 2 ( 1 + 1 2 ) = 3 4 \displaystyle\sum = \tfrac12\Big(1+\tfrac12\Big) = \boxed{\tfrac34} ∑ = 2 1 ( 1 + 2 1 ) = 4 3
Worked example Logs telescope too:∑ n = 2 ∞ ln ( 1 − 1 n 2 ) \displaystyle\sum_{n=2}^\infty \ln\!\Big(1-\tfrac1{n^2}\Big) n = 2 ∑ ∞ ln ( 1 − n 2 1 ) Step 1. 1 − 1 n 2 = ( n − 1 ) ( n + 1 ) n 2 1-\frac1{n^2} = \frac{(n-1)(n+1)}{n^2} 1 − n 2 1 = n 2 ( n − 1 ) ( n + 1 ) = ln ( n − 1 ) + ln ( n + 1 ) − 2 ln n = \ln(n-1)+\ln(n+1)-2\ln n = ln ( n − 1 ) + ln ( n + 1 ) − 2 ln n Why? Logs turn products into sums; this is a "second-difference" of ln n \ln n ln n Step 2 — telescoping regroup. Write the term as a difference of consecutive ratios:
ln ( 1 − 1 n 2 ) = [ ln n + 1 n ] − [ ln n n − 1 ] . \ln\!\Big(1-\tfrac1{n^2}\Big)=\Big[\ln\tfrac{n+1}{n}\Big]-\Big[\ln\tfrac{n}{n-1}\Big]. ln ( 1 − n 2 1 ) = [ ln n n + 1 ] − [ ln n − 1 n ] . Why? Expand: ln n + 1 n − ln n n − 1 = ln ( n + 1 ) + ln ( n − 1 ) − 2 ln n \ln\frac{n+1}{n}-\ln\frac{n}{n-1}=\ln(n+1)+\ln(n-1)-2\ln n ln n n + 1 − ln n − 1 n = ln ( n + 1 ) + ln ( n − 1 ) − 2 ln n b n + 1 − b n b_{n+1}-b_n b n + 1 − b n b n = ln n n − 1 b_n=\ln\frac{n}{n-1} b n = ln n − 1 n Step 3 — partial sum from n = 2 n=2 n = 2 N N N The ln n n − 1 \ln\frac{n}{n-1} ln n − 1 n − ln 2 1 -\ln\frac{2}{1} − ln 1 2 + ln N + 1 N +\ln\frac{N+1}{N} + ln N N + 1 S N = ln N + 1 N − ln 2 1 = ln N + 1 2 N . S_N = \ln\frac{N+1}{N} - \ln\frac{2}{1} = \ln\frac{N+1}{2N}. S N = ln N N + 1 − ln 1 2 = ln 2 N N + 1 . Step 4 — limit. S N = ln N + 1 2 N → N → ∞ ln 1 2 = − ln 2. \displaystyle S_N=\ln\frac{N+1}{2N}\xrightarrow{N\to\infty}\ln\tfrac12 = -\ln 2. S N = ln 2 N N + 1 N → ∞ ln 2 1 = − ln 2.
Common mistake Steel-manning the classic errorsMistake 1 — "interior terms always fully cancel, so the sum is just b 1 b_1 b 1
Why it feels right: In 1 n ( n + 1 ) \frac{1}{n(n+1)} n ( n + 1 ) 1 1 N + 1 → 0 \frac{1}{N+1}\to 0 N + 1 1 → 0 = b 1 =b_1 = b 1 The fix: The sum is b 1 − lim b N + 1 b_1 - \lim b_{N+1} b 1 − lim b N + 1 b n ↛ 0 b_n\not\to 0 b n → 0 ∑ ( n n + 1 − n + 1 n + 2 ) \sum(\frac{n}{n+1}-\frac{n+1}{n+2}) ∑ ( n + 1 n − n + 2 n + 1 ) b 1 = 1 2 b_1=\frac12 b 1 = 2 1 lim b n = 1 \lim b_n=1 lim b n = 1 = 1 2 − 1 = − 1 2 =\frac12-1=-\frac12 = 2 1 − 1 = − 2 1 1 2 \frac12 2 1
Mistake 2 — forgetting extra survivors when the gap > 1.
Why it feels right: "Cancellation happens, so just keep the first and last." But with b n − b n + 2 b_n - b_{n+2} b n − b n + 2 two lead and two trailing terms survive. Always write out 4–5 terms before trusting the pattern.
Mistake 3 — wrong starting index. If a series starts at n = 3 n=3 n = 3 b 3 b_3 b 3 b 1 b_1 b 1
Recall Feynman: explain to a 12-year-oldImagine a line of kids each holding a card. The first kid says "+1", and also "−1 2 \frac12 2 1 1 2 \frac12 2 1 1 3 \frac13 3 1 1 3 \frac13 3 1 1 4 \frac14 4 1 first number (+1, no one before to cancel it) and the very last tiny number survive. As the line gets infinitely long, the last number shrinks to zero — so the whole noisy sum is just 1 . That collapsing-to-the-ends is exactly how a telescope folds shut.
"FIRST minus LAST, the MIDDLE goes past."
S N = b first − b last+1 S_N = b_{\text{first}} - b_{\text{last+1}} S N = b first − b last+1
What defines a telescoping series? A series
∑ a n \sum a_n ∑ a n where
a n = b n − b n + 1 a_n=b_n-b_{n+1} a n = b n − b n + 1 for some sequence
b n b_n b n , so partial sums collapse.
Formula for the partial sum S N S_N S N ∑ n = 1 N ( b n − b n + 1 ) \sum_{n=1}^N (b_n-b_{n+1}) ∑ n = 1 N ( b n − b n + 1 ) S N = b 1 − b N + 1 S_N=b_1-b_{N+1} S N = b 1 − b N + 1 .
Exact value of ∑ n = 1 ∞ 1 n ( n + 1 ) \sum_{n=1}^\infty \frac{1}{n(n+1)} ∑ n = 1 ∞ n ( n + 1 ) 1 1 1 1 , since
S N = 1 − 1 N + 1 → 1 S_N=1-\frac{1}{N+1}\to1 S N = 1 − N + 1 1 → 1 .
What tool usually creates the telescoping difference? Partial fraction decomposition.
∑ n = 1 ∞ 1 n ( n + 2 ) \sum_{n=1}^\infty\frac{1}{n(n+2)} ∑ n = 1 ∞ n ( n + 2 ) 1 3 4 \frac34 4 3 (two terms survive at each end because the gap is 2).
Why isn't the infinite sum always just b 1 b_1 b 1 Because
∑ a n = b 1 − lim n → ∞ b n \sum a_n=b_1-\lim_{n\to\infty}b_n ∑ a n = b 1 − lim n → ∞ b n ; the leftover limit only vanishes if
b n → 0 b_n\to0 b n → 0 .
For a gap-k k k a n = b n − b n + k a_n=b_n-b_{n+k} a n = b n − b n + k k k k leading and
k k k trailing terms.
Condition for a telescoping series to converge? lim n → ∞ b n \lim_{n\to\infty}b_n lim n → ∞ b n must exist (be finite).
Partial Fraction Decomposition — the engine that produces b n − b n + 1 b_n-b_{n+1} b n − b n + 1 Geometric Series — another exact-sum family; contrast with telescoping.Sequence of Partial Sums — telescoping is defined via this sequence collapsing.Convergence Tests for Series — telescoping gives the value , not just convergence.Limits of Sequences — the surviving lim b n \lim b_n lim b n Method of Differences — discrete analogue / same idea in finite sums.
Exact sum, not just convergence
1 over n n+1 = 1/n - 1/n+1
Intuition Hinglish mein samjhoTelescoping series ka idea bilkul simple hai: har term ko ek difference b n − b n + 1 b_n - b_{n+1} b n − b n + 1 b 1 b_1 b 1 b N + 1 b_{N+1} b N + 1 S N = b 1 − b N + 1 S_N = b_1 - b_{N+1} S N = b 1 − b N + 1 lim b n \lim b_n lim b n
Sabse important practical skill hai partial fractions . Jaise 1 n ( n + 1 ) \frac{1}{n(n+1)} n ( n + 1 ) 1 1 n − 1 n + 1 \frac1n - \frac{1}{n+1} n 1 − n + 1 1 = 1 − 1 N + 1 = 1 - \frac{1}{N+1} = 1 − N + 1 1 N → ∞ N\to\infty N → ∞ 1 1 1 exact value milta hai, sirf "converge karti hai" nahi.
Do galtiyon se bacho. Pehli: mat socho ki answer hamesha b 1 b_1 b 1 b 1 − lim b n b_1 - \lim b_n b 1 − lim b n b n b_n b n 2 2 2 b n − b n + 2 b_n - b_{n+2} b n − b n + 2 do-do terms bachte hain, ek nahi. Isliye trust se pehle 4-5 terms haath se likh ke pattern check karo.
Mnemonic yaad rakho: "First minus Last, the Middle goes past." Bas pehla aur aakhri pakdo, aur limit le lo — kaam khatam.