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Question Number 125857 by Dwaipayan Shikari last updated on 14/Dec/20
1+4((1/2))^7 +7(((1.3)/(2.4)))^7 +10(((1.3.5)/(2.4.6)))^7 +...
$$\mathrm{1}+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{7}} +\mathrm{7}\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}\right)^{\mathrm{7}} +\mathrm{10}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\right)^{\mathrm{7}} +… \\ $$
Answered by Olaf last updated on 14/Dec/20
S = 1+Σ_(n=1) ^∞ (1+3n)(Π_(k=1) ^n (((2k−1))/((2k))))^7 S = 1+Σ_(n=1) ^∞ (1+3n)((1/(2^n n!))Π_(k=1) ^n (2k−1))^7 S = 1+Σ_(n=1) ^∞ (1+3n)((1/(2^n n!))Π_(k=1) ^n (((2k−1)(2k))/((2k))))^7 S = 1+Σ_(n=1) ^∞ (1+3n)((((2n)!)/(2^(2n) n!^2 )))^7 Γ(n+(1/2)) = (((2n)!)/(2^(2n) n!))(√π) S = 1+(1/π^(7/2) )Σ_(n=1) ^∞ (1+3n)(((Γ(n+(1/2)))/(Γ(n+1))))^7 ...to be continued...
$$\mathrm{S}\:=\:\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\mathrm{3}{n}\right)\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)}{\left(\mathrm{2}{k}\right)}\right)^{\mathrm{7}} \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\mathrm{3}{n}\right)\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}!}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)\right)^{\mathrm{7}} \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\mathrm{3}{n}\right)\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}!}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}\right)}{\left(\mathrm{2}{k}\right)}\right)^{\mathrm{7}} \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\mathrm{3}{n}\right)\left(\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!^{\mathrm{2}} }\right)^{\mathrm{7}} \\ $$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!}\sqrt{\pi} \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\pi^{\mathrm{7}/\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\mathrm{3}{n}\right)\left(\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\right)^{\mathrm{7}} \\ $$$$…\mathrm{to}\:\mathrm{be}\:\mathrm{continued}… \\ $$
Commented by Dwaipayan Shikari last updated on 14/Dec/20
Great sir!
$${Great}\:{sir}! \\ $$

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