Fie n∈N*.
[tex] \frac{1}{n(n+1)} <\frac{1}{ n^{2} }< \frac{1}{(n-1)n} [/tex]
Astfel: [tex] \frac{1}{2^{2} } < \frac{1}{1*2} \\ \frac{1}{ 3^{2} } < \frac{1}{2*3} \\ .............. \\ \frac{1}{100^{2} } < \frac{1}{99*100} [/tex]
Insumand, obtinem: [tex]a< \frac{1}{1*2}+ \frac{1}{2*3}+...+ \frac{1}{99*100} = \frac{99}{100} ~(SUMA~TELESCOPICA) [/tex]. Asadar [tex] \frac{a}{11}< \frac{ \frac{99}{100} }{11} = \frac{9}{100} => \sqrt{ \frac{a}{11} }< \sqrt{ \frac{9}{100} } =0,3[/tex]......... (1)
Pentru a demonstra "cealalta parte", procedam analog:
[tex] \frac{1}{ 2^{2} }> \frac{1}{2*3} \\ \frac{1}{ 3^{2} }> \frac{1}{3*4} \\............. \\ \frac{1}{ 100^{2} }> \frac{1}{100*101} [/tex]. Insumand, obtinem a>[tex] \frac{1}{2*3}+ \frac{1}{3*4}+...+ \frac{1}{100*101} = \frac{1}{2}- \frac{1}{101}= \frac{99}{202} (SUMA~TELESCOPICA)[/tex]. Deci [tex] \frac{a}{11} > \frac{ \frac{99}{202} }{11}= \frac{9}{202} => \sqrt{ \frac{a}{11}}> \sqrt{ \frac{9}{202}} = \frac{3 \sqrt{202} }{202} > 0,2[/tex]........... (2)
*[tex] \frac{3 \sqrt{202} }{202}>0,2 [/tex] se verifica eventual prin calcul, prin ridicare la patrat sau prin verificarea produsului dintre mezi si extremi.
Din (1) si (2) => [tex]0,2< \sqrt{ \frac{a}{11} } <0.3[/tex]. (Sau cum imi place mie sa spun: [tex] \sqrt{ \frac{a}{11} } [/tex]∈(0,2 ; 0,3).
