prove that (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))<((25)/(36))


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Question Number 213485 by tri26112004 last updated on 06/Nov/24

$p r o v e t h a t$ $\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{2021^{2}} < \frac{25}{36}$
	Answered by mr W last updated on 06/Nov/24

	$w e k n o w :$ $π^{2} < 10$ $1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{2021^{2}} + \frac{1}{2022^{2}} + . . . = \frac{π^{2}}{6}$ $t h e r e f o r e :$ $\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{2021^{2}} = \frac{π^{2}}{6} - 1 - (\frac{1}{2022^{2}} + . . .)$ $< \frac{π^{2}}{6} - 1 = \frac{6 π^{2} - 36}{36}$ $< \frac{60 - 36}{36}$ $< \frac{61 - 36}{36} = \frac{25}{36}$
	Answered by issac last updated on 07/Nov/24
	$prove that (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))<((25)/(36)) Let′s Start with a rigorous proof of Basel Problem Σ_(j=1) ^∞ ((1/j))^2 =(π^2 /6) (Euler′s proof method) .....1) we already know sin(z)=Σ_(j=0) ^∞ (((−)^j )/((2j+1)!))z^(2j+1) sinc(z)=((sin(z))/z) sinc(z)=Σ_(j=0) ^∞ (((−)^j )/((2j+1)!))∙z^(2j) and solution of ((sin(z))/z)=0 is z=nπ , n∈Z\{0} so we can renote ((sin(z))/z)=KΠ_(n∈Z\{0}) (1−(z/(nπ))) K is proportional Const lim_(z→0) ((sin(z))/z)=1 and lim_(z→0) K∙Π_(n=1) ^∞ (1−((z/(nπ)))^2 )=KΠ_(n=1) ^∞ 1=K ∴K=1 so ((sin(z))/z) =Π_(n=1) ^∞ (1−((z/(nπ)))^2 ) ((sin(z))/z) was Σ_(k=0) ^∞ (((−)^k )/((2k+1)!))z^(2k) so Σ_(k=0) ^∞ (((−)^k )/((2k+1)!))z^(2k) =Π_(n=1) ^∞ (1−((z/(nπ)))^2 ) compare the quadratic terms of two equation −(1/6) =Σ_(n=1) ^∞ (−(1/(n^2 π^2 ))) ∴Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) problem (1/2^2 )+(1/3^2 )+.....(1/(2021^2 ))<((25)/(36))...?? (1/1^2 )+(1/2^2 )+....(1/(2021^2 ))<1+((25)/(36)) Σ_(j=1) ^(2021) (1/j^2 )<((61)/(36))≈1.694444....... we alredy know Σ_(j=1) ^(2021) (1/j^2 )<Σ_(j=1) ^∞ (1/j^2 )=(π^2 /6) and (π^2 /6) is approximatly (π^2 /6)≈1.6449340668....... Σ_(j=1) ^(2021) (1/j^2 )<(π^2 /6)<((61)/(36)) Q.E.D$
	$prove that$ $\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{2021^{2}} < \frac{25}{36}$ ${Let}^{'} s Start with a rigorous proof of$ $Basel Problem$ $\underset{j = 1}{\sum^{\infty}} {(\frac{1}{j})}^{2} = \frac{π^{2}}{6} ({Euler}^{'} s proof method)$ $. . . . . 1)$ $we already know \sin (z) = \underset{j = 0}{\sum^{\infty}} \frac{{(-)}^{j}}{(2 j + 1)!} z^{2 j + 1}$ $sinc (z) = \frac{\sin (z)}{z} sinc (z) = \underset{j = 0}{\sum^{\infty}} \frac{{(-)}^{j}}{(2 j + 1)!} \cdot z^{2 j}$ $and solution of \frac{\sin (z)}{z} = 0 is$ $z = n π, n \in Z ∖ {0}$ $so we can renote \frac{\sin (z)}{z} = K \prod_{n \in Z ∖ {0}} (1 - \frac{z}{n π})$ $K is proportional Const$ $\lim_{z \to 0} \frac{\sin (z)}{z} = 1$ $and \lim_{z \to 0} K \cdot \underset{n = 1}{\prod^{\infty}} (1 - {(\frac{z}{n π})}^{2}) = K \underset{n = 1}{\prod^{\infty}} 1 = K$ $∴ K = 1$ $so \frac{\sin (z)}{z} = \underset{n = 1}{\prod^{\infty}} (1 - {(\frac{z}{n π})}^{2})$ $\frac{\sin (z)}{z} was \underset{k = 0}{\sum^{\infty}} \frac{{(-)}^{k}}{(2 k + 1)!} z^{2 k}$ $so \underset{k = 0}{\sum^{\infty}} \frac{{(-)}^{k}}{(2 k + 1)!} z^{2 k} = \underset{n = 1}{\prod^{\infty}} (1 - {(\frac{z}{n π})}^{2})$ $compare the quadratic terms of$ $two equation$ $- \frac{1}{6} = \underset{n = 1}{\sum^{\infty}} (- \frac{1}{n^{2} π^{2}})$ $∴ \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6}$ $problem$ $\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . . . \frac{1}{2021^{2}} < \frac{25}{36} . . . ? ?$ $\frac{1}{1^{2}} + \frac{1}{2^{2}} + . . . . \frac{1}{2021^{2}} < 1 + \frac{25}{36}$ $\underset{j = 1}{\sum^{2021}} \frac{1}{j^{2}} < \frac{61}{36} \approx 1.694444 . . . . . . .$ $we alredy know \underset{j = 1}{\sum^{2021}} \frac{1}{j^{2}} < \underset{j = 1}{\sum^{\infty}} \frac{1}{j^{2}} = \frac{π^{2}}{6}$ $and \frac{π^{2}}{6} is approximatly$ $\frac{π^{2}}{6} \approx 1.6449340668 . . . . . . .$ $\underset{j = 1}{\sum^{2021}} \frac{1}{j^{2}} < \frac{π^{2}}{6} < \frac{61}{36}$ $Q . E . D$
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