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Question Number 80116 by mr W last updated on 31/Jan/20

$$Find$$$$S_m=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\underset{k=1}{\prod ^m}(n+k)}=?$$$$(m⩾2)$$

Commented by mr W last updated on 31/Jan/20

$$someresultsprovedorguessed(?):$$$$S_2=1$$$$S_3=\frac{1}{4}$$$$S_4=\frac{1}{18}(?)$$$$S_5=\frac{1}{96}(?)$$$$S_6=\frac{1}{600}(?)$$$$S_7=\frac{1}{4320}(?)$$$$.....$$$$S_m=\frac{1}{?????intermsofm}$$

Answered by Kamel Kamel last updated on 31/Jan/20

$${\mathrm{S}}_{\mathrm{m}}=\underset{\mathrm{n}=0}{\sum ^{+\mathrm{\infty }}}\frac{\mathrm{n}!}{(\mathrm{m}+\mathrm{n})!}=\frac{1}{\mathrm{\Gamma }\left(\mathrm{m}\right)}\underset{\mathrm{n}=0}{\sum ^{+\mathrm{\infty }}}\frac{\mathrm{\Gamma }(\mathrm{n}+1)\mathrm{\Gamma }\left(\mathrm{m}\right)}{\mathrm{\Gamma }(\mathrm{m}+\mathrm{n}+1)}$$$$=\frac{1}{\mathrm{\Gamma }\left(\mathrm{m}\right)}\underset{\mathrm{n}=0}{\sum ^{+\mathrm{\infty }}}{\int }_0^1{(1-\mathrm{t})}^{\mathrm{m}-1}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}=\frac{1}{\mathrm{\Gamma }\left(\mathrm{m}\right)}{\int }_0^1{(1-\mathrm{t})}^{\mathrm{m}-2}\mathrm{dt}$$$$\therefore {\mathrm{S}}_{\mathrm{m}}=-\frac{1}{(\mathrm{m}-1)\mathrm{\Gamma }\left(\mathrm{m}\right)}{(1-\mathrm{t})}^{\mathrm{m}-1}{\mid }_0^1=\frac{1}{(\mathrm{m}-1)\mathrm{\Gamma }\left(\mathrm{m}\right)}$$$$=\frac{1}{(\mathrm{m}-1)(\mathrm{m}-1)!}$$$$\mathbf{KAMEL}\mathbf{BENAICHA}$$

Commented by mr W last updated on 31/Jan/20

$${it}^{\prime }sperfectsir!i{didn}^{\prime }texpectthat$$$$thereissuchasimpleandeffective$$$$way.thanksalotandcongratulations!$$

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