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Question Number 80116 by mr W last updated on 31/Jan/20
Find S_m =Σ_(n=0) ^∞ (1/(Π_(k=1) ^m (n+k)))=? (m≥2)
$${Find} \\ $$$${S}_{{m}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\left({n}+{k}\right)}=? \\ $$$$\left({m}\geqslant\mathrm{2}\right) \\ $$
Commented by mr W last updated on 31/Jan/20
some results proved or guessed (?): S_2 =1 S_3 =(1/4) S_4 =(1/(18)) (?) S_5 =(1/(96)) (?) S_6 =(1/(600)) (?) S_7 =(1/(4320)) (?) ..... S_m =(1/(????? in terms of m))
$${some}\:{results}\:{proved}\:{or}\:{guessed}\:\left(?\right): \\ $$$${S}_{\mathrm{2}} =\mathrm{1} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{18}}\:\left(?\right) \\ $$$${S}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{96}}\:\left(?\right) \\ $$$${S}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{600}}\:\left(?\right) \\ $$$${S}_{\mathrm{7}} =\frac{\mathrm{1}}{\mathrm{4320}}\:\left(?\right) \\ $$$$….. \\ $$$${S}_{{m}} =\frac{\mathrm{1}}{?????\:{in}\:{terms}\:{of}\:{m}} \\ $$
Answered by Kamel Kamel last updated on 31/Jan/20
S_m =Σ_(n=0) ^(+∞) ((n!)/((m+n)!))=(1/(Γ(m)))Σ_(n=0) ^(+∞) ((Γ(n+1)Γ(m))/(Γ(m+n+1))) =(1/(Γ(m)))Σ_(n=0) ^(+∞) ∫_0 ^1 (1−t)^(m−1) t^n dt=(1/(Γ(m)))∫_0 ^1 (1−t)^(m−2) dt ∴ S_m =−(1/((m−1)Γ(m)))(1−t)^(m−1) ∣_0 ^1 =(1/((m−1)Γ(m))) =(1/((m−1)(m−1)!)) KAMEL BENAICHA
$$\mathrm{S}_{\mathrm{m}} =\underset{\mathrm{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{n}!}{\left(\mathrm{m}+\mathrm{n}\right)!}=\frac{\mathrm{1}}{\Gamma\left(\mathrm{m}\right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)\Gamma\left(\mathrm{m}\right)}{\Gamma\left(\mathrm{m}+\mathrm{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\Gamma\left(\mathrm{m}\right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{m}−\mathrm{1}} \mathrm{t}^{\mathrm{n}} \mathrm{dt}=\frac{\mathrm{1}}{\Gamma\left(\mathrm{m}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{m}−\mathrm{2}} \mathrm{dt} \\ $$$$\:\therefore\:\mathrm{S}_{\mathrm{m}} =−\frac{\mathrm{1}}{\left(\mathrm{m}−\mathrm{1}\right)\Gamma\left(\mathrm{m}\right)}\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{m}−\mathrm{1}} \mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{m}−\mathrm{1}\right)\Gamma\left(\mathrm{m}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{m}−\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{KAMEL}}\:\boldsymbol{\mathrm{BENAICHA}} \\ $$
Commented by mr W last updated on 31/Jan/20
it′s perfect sir! i didn′t expect that there is such a simple and effective way. thanks alot and congratulations!
$${it}'{s}\:{perfect}\:{sir}!\:{i}\:{didn}'{t}\:{expect}\:{that} \\ $$$${there}\:{is}\:{such}\:{a}\:{simple}\:{and}\:{effective} \\ $$$${way}.\:{thanks}\:{alot}\:{and}\:{congratulations}! \\ $$

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