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Question Number 182786 by SANOGO last updated on 14/Dec/22

existant and value m≥1 of S_m =Σ_(n=1 ^ ) ^(+oo) (1/(n(n+1)...(n+m)))

$${existant}\:{and}\:{value}\:{m}\geqslant\mathrm{1}\:{of} \\ $$$${S}_{{m}} =\underset{{n}=\mathrm{1}\:\overset{} {\:}} {\overset{+{oo}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)...\left({n}+{m}\right)}\: \\ $$

Answered by dre23 last updated on 15/Dec/22

S_m =Σ_(n≥1) (1/(n(n+1)..(n+k)...(n+m))) coeficienr of (1/((n+k)))is (((−1)^k )/(k!(m−k)!))=(1/(m!)).(((−1)^k m!)/(k!(m−k)!))=((C_k ^m (−1)^k )/(m!)) S_m =Σ_(n≥1) Σ_(k=0) ^m (((−1)^k C_m ^k )/(m!(n+k))) Σ_(k=0) ^m (−1)^k C_m ^k =(1−1)^m =0 S_m ⇔Σ_(n≥1) Σ_k {{(((−1)^k C_m ^k )/(m!(n+k)))−(((−1)^k C_m ^k )/(m!.n))}+(((−1)^k C_m ^k )/(m!(n+k))) _(=0) } =Σ_(n≥1) {Σ_(k≥0) (((−1)^k C_m ^k )/(m!(n+k)))−(((−1)^k C_m ^k )/(m!n))} =Σ_(k≥0) (((−1)^k C_m ^k )/(m!))Σ_(n≥1) (1/(n+k))−(1/n)....We can change Σ_k Σ_n absolut Cv Σ_(n≥1) (1/(n+k))−(1/n)=Σ_(n≥0) (1/(n+k+1))−(1/(n+1))=Σ_(n≥0) ((−k)/((n+1)(n+k+1))) =−k.((Ψ(k+1)−Ψ(1))/k)=Ψ(1)−Ψ(k+1)=−H_k H_k =Σ_(m=1) ^k (1/m) S_m =Σ_(k=0) ^m (((−1)^(k+1) C_m ^k )/(m!))H_k H_k =∫_0 ^1 ((1−x^k )/(1−x))dx Σ(−1)^(k+1) C_n ^k H_k =∫_0 ^1 ΣC_n ^k (−1)^(k+1) ((1−x^k )/(1−x))dx =∫_0 ^1 ((ΣC_n ^k (−x)^k )/(1−x))dx=∫_0 ^1 (((1−x)^m )/(1−x))dx=∫_0 ^1 (1−x)^(m−1) =(1/m) S_m =(1/(m!)).ΣC_m ^k (−1)^(k+1) H_k =(1/(m!)).(1/m)=(1/(m.m!))

$${S}_{{m}} =\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)..\left({n}+{k}\right)...\left({n}+{m}\right)} \\ $$$${coeficienr}\:{of}\:\frac{\mathrm{1}}{\left({n}+{k}\right)}{is} \\ $$$$\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({m}−{k}\right)!}=\frac{\mathrm{1}}{{m}!}.\frac{\left(−\mathrm{1}\right)^{{k}} {m}!}{{k}!\left({m}−{k}\right)!}=\frac{{C}_{{k}} ^{{m}} \left(−\mathrm{1}\right)^{{k}} }{{m}!} \\ $$$${S}_{{m}} =\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!\left({n}+{k}\right)} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} =\left(\mathrm{1}−\mathrm{1}\right)^{{m}} =\mathrm{0} \\ $$$${S}_{{m}} \Leftrightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}} {\sum}\left\{\left\{\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!\left({n}+{k}\right)}−\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!.{n}}\right\}+\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!\left({n}+{k}\right)}\:_{=\mathrm{0}} \right\} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\left\{\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!\left({n}+{k}\right)}−\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!{n}}\right\} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{m}} ^{{k}} }{{m}!}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}+{k}}−\frac{\mathrm{1}}{{n}}....{We}\:{can}\:{change}\:\underset{{k}} {\sum}\underset{{n}} {\sum} \\ $$$${absolut}\:{Cv} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}+{k}}−\frac{\mathrm{1}}{{n}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}+{k}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{−{k}}{\left({n}+\mathrm{1}\right)\left({n}+{k}+\mathrm{1}\right)} \\ $$$$=−{k}.\frac{\Psi\left({k}+\mathrm{1}\right)−\Psi\left(\mathrm{1}\right)}{{k}}=\Psi\left(\mathrm{1}\right)−\Psi\left({k}+\mathrm{1}\right)=−{H}_{{k}} \\ $$$${H}_{{k}} =\underset{{m}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{m}} \\ $$$${S}_{{m}} =\underset{{k}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {C}_{{m}} ^{{k}} }{{m}!}{H}_{{k}} \\ $$$${H}_{{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{k}} }{\mathrm{1}−{x}}{dx} \\ $$$$\Sigma\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {C}_{{n}} ^{{k}} {H}_{{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \Sigma{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}−{x}^{{k}} }{\mathrm{1}−{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\Sigma{C}_{{n}} ^{{k}} \left(−{x}\right)^{{k}} }{\mathrm{1}−{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{{m}} }{\mathrm{1}−{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{m}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{m}} \\ $$$${S}_{{m}} =\frac{\mathrm{1}}{{m}!}.\Sigma{C}_{{m}} ^{{k}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {H}_{{k}} =\frac{\mathrm{1}}{{m}!}.\frac{\mathrm{1}}{{m}}=\frac{\mathrm{1}}{{m}.{m}!} \\ $$

Commented by SANOGO last updated on 14/Dec/22

merci

$${merci} \\ $$

Commented by dre23 last updated on 16/Dec/22

je vous en prie

$${je}\:{vous}\:{en}\:{prie} \\ $$

Answered by mnjuly1970 last updated on 14/Dec/22

S=(1/m)Σ_(n=1) ^∞ {(1/(n(n+1)...(n+m−1)))−(1/((n+1)(n+2)...(n+m)))} = (1/m) ((1/(1.2.3.4...((m−1))))−0) =(1/(m!))

$$\:\:\:{S}=\frac{\mathrm{1}}{{m}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)...\left({n}+{m}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{m}\right)}\right\} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{m}}\:\left(\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}...\left(\left({m}−\mathrm{1}\right)\right)}−\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}}{{m}!} \\ $$

Commented by SANOGO last updated on 14/Dec/22

merci

$${merci} \\ $$

Commented by dre23 last updated on 15/Dec/22

n=1 we get (1/(mm!))

$${n}=\mathrm{1}\:{we}\:{get}\:\frac{\mathrm{1}}{{mm}!} \\ $$

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