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Question Number 204545 by hardmath last updated on 21/Feb/24
If a = (1/2^2 ) + (1/3^2 ) + ... + (1/(100^2 )) b = 0,99 Prove that: a < b
$$\mathrm{If} \\ $$$$\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} } \\ $$$$\mathrm{b}\:=\:\mathrm{0},\mathrm{99} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{a}\:<\:\mathrm{b} \\ $$
Answered by mr W last updated on 21/Feb/24
a=(1/2^2 )+(1/3^2 )+...+(1/(100^2 )) =(1+(1/2^2 )+(1/3^2 )+...+(1/(100^2 ))+(1/(101^2 ))+...)−1−((1/(101^2 ))+...) <(1+(1/2^2 )+(1/3^2 )+...+(1/(100^2 ))+(1/(101^2 ))+...)−1 =(π^2 /6)−1<0.99
$${a}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{101}^{\mathrm{2}} }+…\right)−\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{101}^{\mathrm{2}} }+…\right) \\ $$$$<\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{101}^{\mathrm{2}} }+…\right)−\mathrm{1} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}<\mathrm{0}.\mathrm{99} \\ $$
Answered by witcher3 last updated on 21/Feb/24
k^2 ≥k(k−1);∀k≥2 ⇒(1/k^2 )≤(1/(k(k−1)))⇒a=Σ_(k=2) ^(100) (1/k^2 )≤Σ_(k=2) ^(100) (1/(k(k−1)))=Σ_(k=2) ^(100) ((1/(k−1))−(1/k))=1−(1/(100))=0.99 a<b
$$\mathrm{k}^{\mathrm{2}} \geqslant\mathrm{k}\left(\mathrm{k}−\mathrm{1}\right);\forall\mathrm{k}\geqslant\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}−\mathrm{1}\right)}\Rightarrow\mathrm{a}=\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{100}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\leqslant\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{100}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}−\mathrm{1}\right)}=\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{100}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{k}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}}=\mathrm{0}.\mathrm{99} \\ $$$$\mathrm{a}<\mathrm{b} \\ $$
Commented by mr W last updated on 21/Feb/24
great approach! but with this method we can not prove (1/2^2 )+(1/3^2 )+...+(1/(100^2 ))<(3/4)
$${great}\:{approach}! \\ $$$${but}\:{with}\:{this}\:{method}\:{we}\:{can}\:{not}\:{prove} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }<\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by witcher3 last updated on 22/Feb/24
exactly same aproch will not Give this but i think That its to prove <(3/4) using interpelation of lagrange approching f(x)=(1/x^2 );withe polynome
$$\mathrm{exactly}\:\mathrm{same}\:\mathrm{aproch}\:\:\mathrm{will}\:\mathrm{not}\:\mathrm{Give}\:\mathrm{this} \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{think}\:\mathrm{That}\:\mathrm{its}\:\mathrm{to}\:\mathrm{prove}\:<\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{using}\:\mathrm{interpelation}\:\mathrm{of}\:\mathrm{lagrange} \\ $$$$\mathrm{approching}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} };\mathrm{withe}\:\mathrm{polynome} \\ $$
Commented by mr W last updated on 22/Feb/24
thanks alot!
$${thanks}\:{alot}! \\ $$
Commented by hardmath last updated on 23/Feb/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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