Question

Aratati ca:1/3^2+1/4^2+1/5^2+.......+1/201^2<1/2

Answer 1

Notez suma cu "S".

$\frac{1}{ 3^{2} } = \frac{1}{3*3}\ \textless \ \frac{1}{2*3}= \frac{1}{2}- \frac{1}{3} \\ \frac{1}{ 4^{2} } = \frac{1}{4*4}\ \textless \ \frac{1}{3*4}= \frac{1}{3}- \frac{1}{4} \\ \frac{1}{ 5^{2} } = \frac{1}{5*5}\ \textless \ \frac{1}{4*5}= \frac{1}{4}- \frac{1}{5} \\ ..................................... \\ \frac{1}{ 201^{2} } = \frac{1}{201*201}\ \textless \ \frac{1}{200*201} = \frac{1}{200}- \frac{1}{201}$

Deci $S\ \textless \ \frac{1}{2}- \frac{1}{3}+ \frac{1}{3}- \frac{1}{4}+ \frac{1}{4}- \frac{1}{5}+...+ \frac{1}{200}- \frac{1}{201}= \\ = \frac{1}{2}- \frac{1}{201}\ \textless \ \frac{1}{2} .$

$=\ \textgreater \ S\ \textless \ \frac{1}{2} .$