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Question Number 55094 by naka3546 last updated on 17/Feb/19

(((1/4) + (1/(16)) + (1/(36)) + (1/(64)) + ...)/(1 + (1/9) + (1/(25)) + (1/(49)) + ...)) = x 3x^2 + 2x − 1 = ?

$$\frac{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\:+\:\frac{\mathrm{1}}{\mathrm{36}}\:+\:\frac{\mathrm{1}}{\mathrm{64}}\:+\:...}{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{25}}\:+\:\frac{\mathrm{1}}{\mathrm{49}}\:+\:...}\:\:=\:\:{x} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}\:\:=\:\:? \\ $$

Commented by maxmathsup by imad last updated on 17/Feb/19

we have (1/4) +(1/(16)) +(1/(16)) +...=1+((1/4)) +((1/4))^2 +....−1 =(1/(1−(1/4))) −1 =(4/3)−1 =(1/3) 1+(1/9) +(1/(25)) +(1/(49)) +...=1+((1/3))^2 +((1/5))^2 +((1/7))^2 +... =Σ_(n=0) ^∞ (1/((2n+1)^2 )) but Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /6) =Σ_(n=1) ^∞ (1/((2n)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒(1−(1/4))Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒(3/4) (π^2 /6) =Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /8) ⇒ x =((1/3)/(π^2 /8)) =((8π^2 )/3) ⇒ 3x^2 +2x−1 =3 (((8π^2 )/3))^2 +2 (((8π^2 )/3))−1 = ((64π^4 )/3) +((16π^2 )/3) −1 .

$${we}\:{have}\:\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+...=\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \:+....−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}\:−\mathrm{1}\:=\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}}\:+\frac{\mathrm{1}}{\mathrm{25}}\:+\frac{\mathrm{1}}{\mathrm{49}}\:+...=\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{2}} +...\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${but}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\Rightarrow\:{x}\:=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\pi^{\mathrm{2}} }{\mathrm{8}}}\:=\frac{\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{1}\:=\mathrm{3}\:\left(\frac{\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{3}}\right)^{\mathrm{2}} \:+\mathrm{2}\:\left(\frac{\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{3}}\right)−\mathrm{1}\:=\:\frac{\mathrm{64}\pi^{\mathrm{4}} }{\mathrm{3}}\:+\frac{\mathrm{16}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{1}\:. \\ $$

Commented by maxmathsup by imad last updated on 17/Feb/19

error from line 6 x =(8/(3π^2 )) ⇒ 3x^2 +2x−1 =3 ((8/(3π^2 )))^2 +2 (8/(3π^2 )) −1 =((64)/(3π^4 )) +((16)/(3π^2 )) −1 .

$${error}\:{from}\:{line}\:\mathrm{6}\:\:\:\:{x}\:=\frac{\mathrm{8}}{\mathrm{3}\pi^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{1}\:=\mathrm{3}\:\left(\frac{\mathrm{8}}{\mathrm{3}\pi^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{2}\:\frac{\mathrm{8}}{\mathrm{3}\pi^{\mathrm{2}} }\:−\mathrm{1}\:=\frac{\mathrm{64}}{\mathrm{3}\pi^{\mathrm{4}} }\:+\frac{\mathrm{16}}{\mathrm{3}\pi^{\mathrm{2}} }\:−\mathrm{1}\:. \\ $$

Answered by Joel578 last updated on 17/Feb/19

Answered by tm888 last updated on 18/Feb/19

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