1 n + 1 + 1 n + 2 + ⋯ + 1 3 n + 1 > 1 {\displaystyle {\frac {1}{n+1}}+{\frac {1}{n+2}}+\cdots +{\frac {1}{3n+1}}>1}
L : 1 n + 1 = 1 4 → 1 1 + 1 + 1 1 + 2 + 1 1 + 3 = 1 1 9 R : 1 1 1 9 > 1 {\displaystyle {\begin{aligned}L:{\frac {1}{n+1}}={\frac {1}{4}}\rightarrow {\frac {1}{1+1}}+{\frac {1}{1+2}}+{\frac {1}{1+3}}=1{\frac {1}{9}}\\R:1\\1{\frac {1}{9}}>1\\\end{aligned}}}
1 k + 1 + 1 k + 2 + ⋯ + 1 3 k + 1 > 1 {\displaystyle {\frac {1}{k+1}}+{\frac {1}{k+2}}+\cdots +{\frac {1}{3k+1}}>1}
1 k + 2 + 1 k + 3 + ⋯ + 1 3 k + 1 ⏟ = ( e n o u g h t o s a y ) 1 − 1 k + 1 + 1 3 k + 2 + 1 3 k + 3 + 1 3 k + 4 > 1 1 − 1 k + 1 + 1 3 k + 2 + 1 3 k + 3 + 1 3 k + 4 > 1 − 1 k + 1 + 1 3 k + 2 + 1 3 k + 3 + 1 3 k + 4 > 0 ↓ − 3 ( 3 k + 2 ) ( 3 k + 4 ) + 3 ( k + 1 ) ( 3 k + 4 ) + ( 3 k + 2 ) ( 3 k + 4 ) + 3 ( k + 1 ) ( 3 k + 2 > 0 − 27 k 2 − 54 k − 24 + 9 k 2 + 21 k + 12 + 9 k 2 + 18 k + 8 + 9 k 2 + 15 k + 10 > 0 6 > 0 {\displaystyle {\begin{aligned}&\underbrace {{\frac {1}{k+2}}+{\frac {1}{k+3}}+\cdots +{\frac {1}{3k+1}}} _{=(enoughtosay)1-{\frac {1}{k+1}}}+{\frac {1}{3k+2}}+{\frac {1}{3k+3}}+{\frac {1}{3k+4}}>1\\&1-{\frac {1}{k+1}}+{\frac {1}{3k+2}}+{\frac {1}{3k+3}}+{\frac {1}{3k+4}}>1\\&-{\frac {1}{k+1}}+{\frac {1}{3k+2}}+{\frac {1}{3k+3}}+{\frac {1}{3k+4}}>0\\&\downarrow \\&-3(3k+2)(3k+4)+3(k+1)(3k+4)+(3k+2)(3k+4)+3(k+1)(3k+2>0\\&-27k^{2}-54k-24+9k^{2}+21k+12+9k^{2}+18k+8+9k^{2}+15k+10>0\\&6>0\\\end{aligned}}}
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טבעי
