Question

Докажите, что $\frac{1}{1^{2} } + \frac{1}{2^{2} } + ... + \frac{1}{100^{2} } \ \textless \ 2$

Answer 1

$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{100^2}<\dfrac{1}{1^2}+\dfrac{1}{1*2}+\dfrac{1}{2*3}+...+\dfrac{1}{99*100}=(*)=1+(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{99}-\dfrac{1}{100})=1+1-\dfrac{1}{100}=2-\dfrac{1}{100}<2$  

Ч.т.д.

$(*)\;\;\;\;\;\;\forall\; n\in N:\;\;\;\;\; \dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{n+1}=> 1=(A+B)n+A =>\left \{ {{A+B=0} \atop {A=1}} \right. =>\left \{ {{A=1} \atop {B=-1}} \right. => \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$