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Question Number 121454 by Dwaipayan Shikari last updated on 08/Nov/20
1+(1/1^2 )+(1/2^2 )+...+(1/((1.2)^2 ))+(1/((2.3)^2 ))+...+(1/((1.2.3)^2 ))+(1/((2.3.4)^2 ))+...+(1/((1.2.3.4)^2 ))+..... Or 1+Σ_(n=1) ^∞ (1/n^2 )+Σ_(n=1) ^∞ (1/((n(n+1))^2 ))+Σ^∞ (1/((n(n+1)(n+2))^2 ))+.....
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}.\mathrm{3}\right)^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}.\mathrm{3}.\mathrm{4}\right)^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)^{\mathrm{2}} }+….. \\ $$$$ \\ $$$${Or} \\ $$$$\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} }+\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right)^{\mathrm{2}} }+….. \\ $$
Commented by Dwaipayan Shikari last updated on 08/Nov/20
I think all terms should be Π_(n=1) ^∞ (1+(1/n^2 ))
$${I}\:{think}\:{all}\:{terms}\:{should}\:{be}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$
Commented by mindispower last updated on 09/Nov/20
((sin(x))/x)=Π_(n≥1) (1−(x^2 /(n^2 π^2 ))) x=i⇒((sin(iπ))/(iπ))=Π(1+(1/(n^2 π^2 )))=((sh(π))/π) ln(Π_(n≥1) (1+(1/n^2 )))=Σ_(n≥1) ln(1+(1/n^2 ))=Σ.Σ_(k≥1) (((−1)^k )/(kn^(2k) )) =Σ_(k≥1) (((−1)^k )/k)Σ_(n≥1) (1/n^(2k) )=Σ_(k≥1) (((−1)^k )/k)ζ(2k) we[get nice identitie Σ(((−1)^k ζ(2k))/k)=((sh(π))/π)
$$\frac{{sin}\left({x}\right)}{{x}}=\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$$${x}={i}\Rightarrow\frac{{sin}\left({i}\pi\right)}{{i}\pi}=\Pi\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right)=\frac{{sh}\left(\pi\right)}{\pi} \\ $$$${ln}\left(\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\Sigma.\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{kn}^{\mathrm{2}{k}} } \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}{k}} }=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\zeta\left(\mathrm{2}{k}\right) \\ $$$${we}\left[{get}\:{nice}\:{identitie}\right. \\ $$$$\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} \zeta\left(\mathrm{2}{k}\right)}{{k}}=\frac{{sh}\left(\pi\right)}{\pi} \\ $$$$ \\ $$
Answered by Olaf last updated on 08/Nov/20
S = Σ_(k=0) ^∞ Σ_(n=1) ^∞ (1/(Π_(p=n) ^(n+k) p^2 )) S = Σ_(k=0) ^∞ Σ_(n=1) ^∞ (((n−1)!^2 )/((n+k)!^2 )) S = Σ_(n=1) ^∞ Σ_(k=0) ^∞ (((n−1)!^2 )/((n+k)!^2 )) S = Σ_(n=1) ^∞ (n−1)!^2 Σ_(k=0) ^∞ (1/((n+k)!^2 )) ...see Bessel functions
$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\underset{{p}={n}} {\overset{{n}+{k}} {\prod}}{p}^{\mathrm{2}} } \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}−\mathrm{1}\right)!^{\mathrm{2}} }{\left({n}+{k}\right)!^{\mathrm{2}} } \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}−\mathrm{1}\right)!^{\mathrm{2}} }{\left({n}+{k}\right)!^{\mathrm{2}} } \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right)!^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+{k}\right)!^{\mathrm{2}} } \\ $$$$…\mathrm{see}\:\mathrm{Bessel}\:\mathrm{functions}\: \\ $$

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