Problem
ALG-B1-M05-P016 Estimate of a Telescoping Sum
#16
★★★★★ Level 5 of 5
Prove that \(\sum_{k=1}^{n}\frac1{k^2}<2\) for all \(n\).
For \(k\ge2\), compare \(\frac1{k^2}\) with \(\frac1{k-1}-\frac1k\).
For \(k\ge2\), \(k^2>k(k-1)\), so \(\frac1{k^2}<\frac1{k(k-1)}=\frac1{k-1}-\frac1k\).
Thus the sum is less than \(1+\sum_{k=2}^{n}\left(\frac1{k-1}-\frac1k\right)=1+1-\frac1n<2\).
Estimation by telescoping.