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Question Number 99803 by student work last updated on 23/Jun/20

Π_(k=1) ^∞ (1+(1/k^2 ))=? helpe me

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)=? \\ $$$$\mathrm{helpe}\:\mathrm{me} \\ $$

Answered by smridha last updated on 23/Jun/20

we know sin(z)=zΠ_(k=1) ^∞ [1−((z/(k𝛑)))^2 ] so Π_(k=1) ^∞ (1+(1/k^2 ))=lim_(z→i𝛑) ((sin(z))/z) =lim_(z→i𝛑) ((e^(iz) −e^(−iz) )/(2iz)) =((e^(−𝛑) −e^𝛑 )/(−2𝛑))=(1/𝛑)sinh(𝛑)≈3.676....

$$\boldsymbol{{we}}\:\boldsymbol{{know}}\:\boldsymbol{{sin}}\left(\boldsymbol{{z}}\right)=\boldsymbol{{z}}\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\infty} {\prod}}\left[\mathrm{1}−\left(\frac{\boldsymbol{{z}}}{\boldsymbol{{k}\pi}}\right)^{\mathrm{2}} \right] \\ $$$$\:\boldsymbol{{so}}\:\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{k}}^{\mathrm{2}} }\right)=\underset{\boldsymbol{{z}}\rightarrow\boldsymbol{{i}\pi}} {\mathrm{lim}}\frac{\boldsymbol{{sin}}\left(\boldsymbol{{z}}\right)}{\boldsymbol{{z}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\boldsymbol{{z}}\rightarrow\boldsymbol{{i}\pi}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} −\boldsymbol{{e}}^{−\boldsymbol{{iz}}} }{\mathrm{2}\boldsymbol{{iz}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{e}}^{−\boldsymbol{\pi}} −\boldsymbol{{e}}^{\boldsymbol{\pi}} }{−\mathrm{2}\boldsymbol{\pi}}=\frac{\mathrm{1}}{\boldsymbol{\pi}}\boldsymbol{{sinh}}\left(\boldsymbol{\pi}\right)\approx\mathrm{3}.\mathrm{676}.... \\ $$

Answered by mathmax by abdo last updated on 23/Jun/20

we have sin(πz) =πz Π_(n=1) ^∞ (1−(z^2 /n^2 )) z =i ⇒sin(iπ) =iπ Π_(n=1) ^∞ (1+(1/n^2 )) ⇒Π_(n=1) ^∞ (1+(1/n^2 )) =((sin(iπ))/(iπ)) =((e^(−π) −e^π )/(2i×(iπ))) =((e^π −e^(−π) )/(2 ))×(1/π) =((sh(π))/π)

$$\mathrm{we}\:\mathrm{have}\:\mathrm{sin}\left(\pi\mathrm{z}\right)\:=\pi\mathrm{z}\:\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{z}\:=\mathrm{i}\:\Rightarrow\mathrm{sin}\left(\mathrm{i}\pi\right)\:=\mathrm{i}\pi\:\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\:\Rightarrow\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\:=\frac{\mathrm{sin}\left(\mathrm{i}\pi\right)}{\mathrm{i}\pi} \\ $$$$=\frac{\mathrm{e}^{−\pi} −\mathrm{e}^{\pi} }{\mathrm{2i}×\left(\mathrm{i}\pi\right)}\:=\frac{\mathrm{e}^{\pi} −\mathrm{e}^{−\pi} }{\mathrm{2}\:}×\frac{\mathrm{1}}{\pi}\:=\frac{\mathrm{sh}\left(\pi\right)}{\pi} \\ $$$$ \\ $$

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