Question

1/2²+1/3²+...+1/2011² mai mare decat 2010/2011​

Answer 1

$\displaystyle\bf\\\boxed{\bf\frac{1}{k^2} < \frac{1}{k(k-1)},~\forall~k~pozitiv~nenul}.\\\\demonstratie~:~\frac{1}{k^2} < \frac{1}{k(k-1)} \Leftrightarrow k^2>k(k-1) \Leftrightarrow k^2>k^2-k \Leftrightarrow k^2-k^2+k>0 \Leftrightarrow\\k>0.\\\frac{1}{2^2} < \frac{1}{1\cdot2}.\\\frac{1}{3^2} < \frac{1}{2\cdot3}.\\.\\.\\.\\\frac{1}{2011^2} < \frac{1}{2010\cdot2011}.\\----------(+)---\\\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2} < \frac{1}{2} + \frac{1}{2\cdot3}+...+\frac{1}{2010\cdot2011}.$

$\displaystyle\bf\\\frac{1}{k(k+1)} = \frac{A}{k} - \frac{B}{k+1} =\frac{A(k+1)-Bk}{k(k+1)} \Leftrightarrow\\ Ak+A-Bk=1 \Leftrightarrow k(A-B)+A=1,~daca~A=B=1,~se~verifica.\\deci,~\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.\\notam~S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}.\\S<\frac{1}{2} + \frac{1}{2\cdot3}+...+\frac{1}{2010\cdot2011}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010} - \frac{1}{2011} =\\\\1-\frac{1}{2011}=\frac{2010}{2011},~\boxed{\bf Q.E.D}~.$