Compute ∑i=1n(1i−1i+1)\sum_{i=1}^{n} \left(\frac{1}{i} - \frac{1}{i+1}\right)∑i=1n(i1−i+11). Expand: (11−12)+(12−13)+(13−14)+...+(1n−1n+1)\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + ... + \left(\frac{1}{n} - \frac{1}{n+1}\right)(11−21)+(21−31)+(31−41)+...+(n1−n+11). The −12-\frac{1}{2}−21 and +12+\frac{1}{2}+21 cancel, the −13-\frac{1}{3}−31 and +13+\frac{1}{3}+31 cancel, and so on. You're left with 11−1n+1=1−1n+1\frac{1}{1} - \frac{1}{n+1} = 1 - \frac{1}{n+1}11−n+11=1−n+11.