Diketahui P(n) : 1+122+132+142+⋯+1n2≤2−1nP\left(n\right)\ :\ 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots+\frac{1}{n^2}\le2-\frac{1}{n}P(n) : 1+221+321+421+⋯+n21≤2−n1 untuk setiap bilangan asli nnn. Diandaikan benar untuk n=kn=kn=k, maka akan dibuktikan benar bahwa ....
1+122+132+142+...+1n2≤2−1k1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\le2-\frac{1}{k}1+221+321+421+...+n21≤2−k1
1+122+132+142+...+1n2≤2−1k+11+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\le2-\frac{1}{k+1}1+221+321+421+...+n21≤2−k+11
1+122+132+142+...+1k2≤2−1k1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{k^2}\le2-\frac{1}{k}1+221+321+421+...+k21≤2−k1
1+122+132+142+...+1k2+1(k+1)2≤2−1k+11+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le2-\frac{1}{k+1}1+221+321+421+...+k21+(k+1)21≤2−k+11
1+122+132+142+...+1k2+1(k+1)2≤2−1k1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le2-\frac{1}{k}1+221+321+421+...+k21+(k+1)21≤2−k1