Toolkit 5Telescoping series11⋅2+12⋅3+⋯+1n(n+1)=(1−12)+⋯+(1n−1n+1)=1−1n+1\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)}=\left(1-\frac{1}{2}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}1⋅21+2⋅31+⋯+n(n+1)1=(1−21)+⋯+(n1−n+11)=1−n+11ProofFor every positive integer kkk,1k(k+1)=1k−1k+1.\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.k(k+1)1=k1−k+11.Thus,11⋅2=1−12,\frac{1}{1\cdot 2} = 1 - \frac{1}{2},1⋅21=1−21,12⋅3=12−13,\frac{1}{2\cdot 3} = \frac{1}{2} - \frac{1}{3},2⋅31=21−31,⋮\vdots⋮1n(n+1)=1n−1n+1.\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.n(n+1)1=n1−n+11.Adding these identities, all intermediate terms cancel. Therefore,11⋅2+12⋅3+⋯+1n(n+1)=1−1n+1.□\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}. \quad\square1⋅21+2⋅31+⋯+n(n+1)1=1−n+11.□Related ProblemsCoreAMC 10B 2022 (Problem 9)AMC 8 2022 (Problem 8)AMC 12B 2025 (Problem 7)MATHCOUNTS 2026 State Sprint Round (Problem 24)AIME II 2025 (Problem 4)