[tex] \frac{1}{k^{2} } = \frac{1}{k*k} < \frac{1}{(k-1)k} [/tex]
Adica: [tex] \frac{1}{ 2^{2} } < \frac{1}{1*2} \\ \frac{1}{ 3^{2} } < \frac{1}{2*3} \\ .............. \\ \frac{1}{2011 ^{2} } < \frac{1}{2010*2011} [/tex]
Le aduni pe toate si obtii:
[tex] \frac{1}{2^{2} } + \frac{1}{ 3^{2} } +...+ \frac{1}{ 2011^{2} } < \frac{1}{1*2}+ \frac{1}{2*3}+...+ \frac{1}{2010*2011} = 1- \frac{1}{2}+ \frac{1}{2} - \frac{1}{3} [/tex] [tex]+...+ \frac{1}{2010}- \frac{1}{2011}= 1- \frac{1}{2011} = \frac{2010}{2011} [/tex]
Deci suma este mai mica decat [tex] \frac{2010}{2011} [/tex].
