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Question Number 213485 by tri26112004 last updated on 06/Nov/24
prove that (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))<((25)/(36))
$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }<\frac{\mathrm{25}}{\mathrm{36}} \\ $$
Answered by mr W last updated on 06/Nov/24
we know: π^2 <10 1+(1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))+(1/(2022^2 ))+...=(π^2 /6) therefore: (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))=(π^2 /6)−1−((1/(2022^2 ))+...) <(π^2 /6)−1=((6π^2 −36)/(36)) <((60−36)/(36)) <((61−36)/(36))=((25)/(36))
$${we}\:{know}: \\ $$$$\pi^{\mathrm{2}} <\mathrm{10} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} }+…=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${therefore}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} }+…\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}=\frac{\mathrm{6}\pi^{\mathrm{2}} −\mathrm{36}}{\mathrm{36}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\frac{\mathrm{60}−\mathrm{36}}{\mathrm{36}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\frac{\mathrm{61}−\mathrm{36}}{\mathrm{36}}=\frac{\mathrm{25}}{\mathrm{36}} \\ $$
Answered by issac last updated on 07/Nov/24
prove that (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))<((25)/(36)) Let′s Start with a rigorous proof of Basel Problem Σ_(j=1) ^∞ ((1/j))^2 =(π^2 /6) (Euler′s proof method) .....1) we already know sin(z)=Σ_(j=0) ^∞ (((−)^j )/((2j+1)!))z^(2j+1) sinc(z)=((sin(z))/z) sinc(z)=Σ_(j=0) ^∞ (((−)^j )/((2j+1)!))∙z^(2j) and solution of ((sin(z))/z)=0 is z=nπ , n∈Z\{0} so we can renote ((sin(z))/z)=KΠ_(n∈Z\{0}) (1−(z/(nπ))) K is proportional Const lim_(z→0) ((sin(z))/z)=1 and lim_(z→0) K∙Π_(n=1) ^∞ (1−((z/(nπ)))^2 )=KΠ_(n=1) ^∞ 1=K ∴K=1 so ((sin(z))/z) =Π_(n=1) ^∞ (1−((z/(nπ)))^2 ) ((sin(z))/z) was Σ_(k=0) ^∞ (((−)^k )/((2k+1)!))z^(2k) so Σ_(k=0) ^∞ (((−)^k )/((2k+1)!))z^(2k) =Π_(n=1) ^∞ (1−((z/(nπ)))^2 ) compare the quadratic terms of two equation −(1/6) =Σ_(n=1) ^∞ (−(1/(n^2 π^2 ))) ∴Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) problem (1/2^2 )+(1/3^2 )+.....(1/(2021^2 ))<((25)/(36))...?? (1/1^2 )+(1/2^2 )+....(1/(2021^2 ))<1+((25)/(36)) Σ_(j=1) ^(2021) (1/j^2 )<((61)/(36))≈1.694444....... we alredy know Σ_(j=1) ^(2021) (1/j^2 )<Σ_(j=1) ^∞ (1/j^2 )=(π^2 /6) and (π^2 /6) is approximatly (π^2 /6)≈1.6449340668....... Σ_(j=1) ^(2021) (1/j^2 )<(π^2 /6)<((61)/(36)) Q.E.D
$$\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }<\frac{\mathrm{25}}{\mathrm{36}} \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Start}\:\mathrm{with}\:\mathrm{a}\:\mathrm{rigorous}\:\mathrm{proof}\:\mathrm{of} \\ $$$$\mathrm{Basel}\:\mathrm{Problem}\: \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{{j}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\left(\mathrm{Euler}'\mathrm{s}\:\mathrm{proof}\:\mathrm{method}\right) \\ $$$$\left……\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{already}\:\mathrm{know}\:\mathrm{sin}\left({z}\right)=\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\right)^{{j}} }{\left(\mathrm{2}{j}+\mathrm{1}\right)!}{z}^{\mathrm{2}{j}+\mathrm{1}} \\ $$$$\mathrm{sinc}\left({z}\right)=\frac{\mathrm{sin}\left({z}\right)}{{z}}\:\:\mathrm{sinc}\left({z}\right)=\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\right)^{{j}} }{\left(\mathrm{2}{j}+\mathrm{1}\right)!}\centerdot{z}^{\mathrm{2}{j}} \:\:\: \\ $$$$\mathrm{and}\:\:\mathrm{solution}\:\mathrm{of}\:\:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}=\mathrm{0}\:\mathrm{is} \\ $$$${z}={n}\pi\:,\:{n}\in\mathbb{Z}\backslash\left\{\mathrm{0}\right\} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}={K}\underset{{n}\in\mathbb{Z}\backslash\left\{\mathrm{0}\right\}} {\prod}\left(\mathrm{1}−\frac{{z}}{{n}\pi}\right) \\ $$$${K}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{Const} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\left({z}\right)}{{z}}=\mathrm{1}\: \\ $$$$\mathrm{and}\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{K}\centerdot\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\mathrm{1}−\left(\frac{{z}}{{n}\pi}\right)^{\mathrm{2}} \right)={K}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\mathrm{1}={K} \\ $$$$\therefore{K}=\mathrm{1} \\ $$$$\mathrm{so}\:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\mathrm{1}−\left(\frac{{z}}{{n}\pi}\right)^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{sin}\left({z}\right)}{{z}}\:\mathrm{was}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}{z}^{\mathrm{2}{k}} \\ $$$$\mathrm{so}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}{z}^{\mathrm{2}{k}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\mathrm{1}−\left(\frac{{z}}{{n}\pi}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{compare}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{terms}\:\mathrm{of}\: \\ $$$$\mathrm{two}\:\mathrm{equation} \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(−\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$$$\therefore\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{problem} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…..\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }<\frac{\mathrm{25}}{\mathrm{36}}…?? \\ $$$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+….\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }<\mathrm{1}+\frac{\mathrm{25}}{\mathrm{36}} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\:\frac{\mathrm{1}}{{j}^{\mathrm{2}} }<\frac{\mathrm{61}}{\mathrm{36}}\approx\mathrm{1}.\mathrm{694444}……. \\ $$$$\mathrm{we}\:\mathrm{alredy}\:\mathrm{know}\:\underset{{j}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\:\frac{\mathrm{1}}{{j}^{\mathrm{2}} }<\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{j}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{and}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\mathrm{is}\:\mathrm{approximatly}\: \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\approx\mathrm{1}.\mathrm{6449340668}……. \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\:\frac{\mathrm{1}}{{j}^{\mathrm{2}} }<\frac{\pi^{\mathrm{2}} }{\mathrm{6}}<\frac{\mathrm{61}}{\mathrm{36}} \\ $$$$\mathbb{Q}.\mathbb{E}.\mathbb{D} \\ $$

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