Notez cu [tex]S_n[/tex] distanța parcursă în n secunde.
Aceasta va fi:
[tex]S_n=\dfrac{1}{10}+\dfrac{1}{10\cdot2^2}+\dfrac{1}{10\cdot3^2}+\ldots+\dfrac{1}{10\cdot n^2}=[/tex]
[tex]=\dfrac{1}{10}\cdot(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{n^2})<[/tex]
[tex]\ < \dfrac{1}{10}\cdot(1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\ldots+\dfrac{1}{(n-1)n})=[/tex]
[tex]=\dfrac{1}{10}\cdot(1+\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\ldots+\dfrac{1}{n-1}-\dfrac1n)=[/tex]
[tex]=\dfrac{1}{10}(2-\dfrac1n)[/tex]
[tex] \lim_{n \to \infty} S_n =\dfrac{2}{10}=0,2<0,21[/tex]
