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Question Number 5921 by FilupSmith last updated on 05/Jun/16

Does: (1) Σ_(i=1) ^∞ Γ((1/i)) converge? (2) Π_(i=1) ^∞ Γ((1/i)) converge?

$$\mathrm{Does}:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\Gamma\left(\frac{\mathrm{1}}{{i}}\right)\:\:\mathrm{converge}? \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{i}}\right)\:\:\mathrm{converge}? \\ $$

Commented by Yozzii last updated on 05/Jun/16

(1) Γ(x)Γ(1−x)=(π/(sinπx)) Γ(i^(−1) )=(π/(Γ(1−i^(−1) )sin(π/i))) Γ(x)=∫_0 ^∞ u^(x−1) e^(−t) dt⇒Γ(1)=∫_0 ^∞ e^(−t) dt=1 lim_(i→∞) Γ(i^(−1) )=(π/(Γ(1)sinπ×0))=(π/(Γ(1)×0))=(π/(1×0))=(π/0)≠0 Since lim_(i→∞) Γ(i^(−1) )≠0,Σ_(i=1) ^∞ Γ(i^(−1) ) is not convergent. (ii) Let (a_n ) be a sequence with each a_n >0. If Π_(n=1) ^∞ a_n converges to a limit l and l≠0, then (a_n ) converges and lim_(n→∞) a_n =1. ⇒If (a_n ) diverges, so that lim_(n→∞) a_n ≠1, then Π_(n=1) ^∞ a_n diverges also. Since lim_(i→∞) Γ(i^(−1) ) does not exist and Γ(i^(−1) )>0 for i∈N, then Π_(i=1) ^∞ Γ(i^(−1) ) diverges.

$$\left(\mathrm{1}\right)\:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\pi{x}} \\ $$$$\Gamma\left({i}^{−\mathrm{1}} \right)=\frac{\pi}{\Gamma\left(\mathrm{1}−{i}^{−\mathrm{1}} \right){sin}\frac{\pi}{{i}}} \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {u}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\Rightarrow\Gamma\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {dt}=\mathrm{1} \\ $$$$\underset{{i}\rightarrow\infty} {\mathrm{lim}}\Gamma\left({i}^{−\mathrm{1}} \right)=\frac{\pi}{\Gamma\left(\mathrm{1}\right){sin}\pi×\mathrm{0}}=\frac{\pi}{\Gamma\left(\mathrm{1}\right)×\mathrm{0}}=\frac{\pi}{\mathrm{1}×\mathrm{0}}=\frac{\pi}{\mathrm{0}}\neq\mathrm{0} \\ $$$${Since}\:\underset{{i}\rightarrow\infty} {\mathrm{lim}}\Gamma\left({i}^{−\mathrm{1}} \right)\neq\mathrm{0},\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\Gamma\left({i}^{−\mathrm{1}} \right)\:{is}\:{not}\:{convergent}. \\ $$$$\left({ii}\right)\:{Let}\:\left({a}_{{n}} \right)\:{be}\:{a}\:{sequence}\:{with}\:{each} \\ $$$${a}_{{n}} >\mathrm{0}.\:{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}{a}_{{n}} \:{converges}\:{to}\:{a}\:{limit} \\ $$$${l}\:{and}\:{l}\neq\mathrm{0},\:{then}\:\left({a}_{{n}} \right)\:{converges}\:{and}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\mathrm{1}. \\ $$$$\Rightarrow{If}\:\left({a}_{{n}} \right)\:{diverges},\:{so}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{1}, \\ $$$${then}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}{a}_{{n}} \:{diverges}\:{also}. \\ $$$${Since}\:\underset{{i}\rightarrow\infty} {\mathrm{lim}}\Gamma\left({i}^{−\mathrm{1}} \right)\:{does}\:{not}\:{exist}\:{and}\:\Gamma\left({i}^{−\mathrm{1}} \right)>\mathrm{0}\:{for}\:{i}\in\mathbb{N}, \\ $$$${then}\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\Gamma\left({i}^{−\mathrm{1}} \right)\:{diverges}. \\ $$$$ \\ $$

Commented by FilupSmith last updated on 05/Jun/16

Amazing!

$$\mathrm{Amazing}! \\ $$

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