הוכחה באינדוציה (או בכל דרך אחרת), כי עבור כל n {\displaystyle \ n} טבעי גדול מ- 1 {\displaystyle \ 1} מתקיים : 1 2 2 + 1 3 2 + ⋯ + 1 n 2 < n − 1 n {\displaystyle {\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots +{\frac {1}{n^{2}}}<{\frac {n-1}{n}}}
1 2 2 + 1 3 2 + ⋯ + 1 n 2 < n − 1 n {\displaystyle {\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots +{\frac {1}{n^{2}}}<{\frac {n-1}{n}}}
L : 1 n 2 = 1 2 2 = 1 R : 2 − 1 2 = 1 2 1 4 < 1 2 √ {\displaystyle {\begin{aligned}&L:{\frac {1}{n^{2}}}={\frac {1}{2^{2}}}=1\\&R:{\frac {2-1}{2}}={\frac {1}{2}}\\&{\frac {1}{4}}<{\frac {1}{2}}\surd \\\end{aligned}}}
1 2 2 + 1 3 2 + ⋯ + 1 k 2 < k − 1 k {\displaystyle {\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots +{\frac {1}{k^{2}}}<{\frac {k-1}{k}}}
1 2 2 + 1 3 2 + ⋯ + 1 k 2 + 1 ( k + 1 ) 2 < k k + 1 1 2 2 + 1 3 2 + ⋯ + 1 k 2 ⏟ k − 1 k + 1 ( k + 1 ) 2 < k k + 1 k − 1 k + 1 ( k + 1 ) 2 < k k + 1 ( k − 1 ) ( k + 1 ) 2 + k < k 2 ( k + 1 ) ( k − 1 ) ( k 2 + 2 k + 1 ) + k < k 3 + k 2 k 3 + 2 k 2 + k − k 2 − 2 k − 1 + k < k 3 + k 2 − 1 < 0 {\displaystyle {\begin{aligned}&{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots +{\frac {1}{k^{2}}}+{\frac {1}{(k+1)^{2}}}<{\frac {k}{k+1}}\\&\underbrace {{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots +{\frac {1}{k^{2}}}} _{\frac {k-1}{k}}+{\frac {1}{(k+1)^{2}}}<{\frac {k}{k+1}}\\&{\frac {k-1}{k}}+{\frac {1}{(k+1)^{2}}}<{\frac {k}{k+1}}\\&(k-1)(k+1)^{2}+k<k^{2}(k+1)\\&(k-1)(k^{2}+2k+1)+k<k^{3}+k^{2}\\&k^{3}+2k^{2}+k-k^{2}-2k-1+k<k^{3}+k^{2}\\&-1<0\\\end{aligned}}}
האינדוקציה נכונה על פי 4 שלבי האינדוקציה.
[עריכה]
טבעי[עריכה]
