4 3 ⋅ 9 8 ⋅ 16 15 ⋯ 2019 2 ( 2019 − 1 ) ( 2019 + 1 ) < 2019 n 2 . \frac43\cdot\frac98\cdot\frac{16}{15}\cdots \frac{2019^2}{(2019-1)(2019+1)}<\frac{2019}{n^2}. 3 4 ⋅ 8 9 ⋅ 15 16 ⋯ ( 2019 − 1 ) ( 2019 + 1 ) 201 9 2 < n 2 2019 . Svaki faktor na lijevoj strani nejednakosti možemo zapisati u obliku:
x 2 x 2 − 1 = x ⋅ x ( x − 1 ) ( x + 1 ) = x x − 1 ⋅ x x + 1 . \frac{x^2}{x^2-1}=\frac{x\cdot x}{(x-1)(x+1)} =\frac{x}{x-1}\cdot\frac{x}{x+1}. x 2 − 1 x 2 = ( x − 1 ) ( x + 1 ) x ⋅ x = x − 1 x ⋅ x + 1 x . Slijedi redom
2 ⋅ 2 1 ⋅ 3 ⋅ 3 ⋅ 3 2 ⋅ 4 ⋅ 4 ⋅ 4 3 ⋅ 5 ⋅ 5 ⋅ 5 4 ⋅ 6 ⋯ 2019 ⋅ 2019 2018 ⋅ 2020 < 2019 n 2 \frac{2\cdot2}{1\cdot3}\cdot \frac{3\cdot3}{2\cdot4}\cdot \frac{4\cdot4}{3\cdot5}\cdot \frac{5\cdot5}{4\cdot6}\cdots \frac{2019\cdot2019}{2018\cdot2020} <\frac{2019}{n^2} 1 ⋅ 3 2 ⋅ 2 ⋅ 2 ⋅ 4 3 ⋅ 3 ⋅ 3 ⋅ 5 4 ⋅ 4 ⋅ 4 ⋅ 6 5 ⋅ 5 ⋯ 2018 ⋅ 2020 2019 ⋅ 2019 < n 2 2019 2 1 ⋅ ( 2 3 ⋅ 3 2 ) ⋅ ( 3 4 ⋅ 4 3 ) ⋯ ( 2018 2019 ⋅ 2019 2018 ) ⋅ 2019 2020 < 2019 n 2 . \frac21\cdot \left(\frac23\cdot\frac32\right)\cdot \left(\frac34\cdot\frac43\right)\cdots \left(\frac{2018}{2019}\cdot\frac{2019}{2018}\right)\cdot \frac{2019}{2020}<\frac{2019}{n^2}. 1 2 ⋅ ( 3 2 ⋅ 2 3 ) ⋅ ( 4 3 ⋅ 3 4 ) ⋯ ( 2019 2018 ⋅ 2018 2019 ) ⋅ 2020 2019 < n 2 2019 . Uočimo da su svaka dva susjedna faktora počevši od drugog po redu do pretposljednjeg, međusobno recipročni pa je njihov umnožak jednak 1. Nakon toga dobivamo
2 ⋅ 2019 2020 < 2019 n 2 , \frac{2\cdot2019}{2020}<\frac{2019}{n^2}, 2020 2 ⋅ 2019 < n 2 2019 , što vrijedi ako i samo ako je n 2 < 1010 n^2<1010 n 2 < 1010 n ≠ 0 n\neq0 n = 0
0 < ∣ n ∣ < 1010 , odnosno − 1010 < n < 1010 , n ≠ 0. 0<|n|<\sqrt{1010}, \qquad\text{odnosno}\qquad -\sqrt{1010}<n<\sqrt{1010},\ n\neq0. 0 < ∣ n ∣ < 1010 , odnosno − 1010 < n < 1010 , n = 0. Konačno, traženi cijeli brojevi n n n { − 31 , − 30 , … , − 1 , 1 , 2 , … , 30 , 31 } \{-31,-30,\ldots,-1,1,2,\ldots,30,31\} { − 31 , − 30 , … , − 1 , 1 , 2 , … , 30 , 31 }
