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


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 182786 by SANOGO last updated on 14/Dec/22

$e x i s t a n t a n d v a l u e m ⩾ 1 o f$ $S_{m} = \underset{n = 1 \overset{}{}}{\sum^{+ o o}} \frac{1}{n (n + 1) . . . (n + m)}$
	Answered by dre23 last updated on 15/Dec/22
	$S_m =Σ_(n≥1) (1/(n(n+1)..(n+k)...(n+m))) coeficienr of (1/((n+k)))is (((−1)^k )/(k!(m−k)!))=(1/(m!)).(((−1)^k m!)/(k!(m−k)!))=((C_k ^m (−1)^k )/(m!)) S_m =Σ_(n≥1) Σ_(k=0) ^m (((−1)^k C_m ^k )/(m!(n+k))) Σ_(k=0) ^m (−1)^k C_m ^k =(1−1)^m =0 S_m ⇔Σ_(n≥1) Σ_k {{(((−1)^k C_m ^k )/(m!(n+k)))−(((−1)^k C_m ^k )/(m!.n))}+(((−1)^k C_m ^k )/(m!(n+k))) _(=0) } =Σ_(n≥1) {Σ_(k≥0) (((−1)^k C_m ^k )/(m!(n+k)))−(((−1)^k C_m ^k )/(m!n))} =Σ_(k≥0) (((−1)^k C_m ^k )/(m!))Σ_(n≥1) (1/(n+k))−(1/n)....We can change Σ_k Σ_n absolut Cv Σ_(n≥1) (1/(n+k))−(1/n)=Σ_(n≥0) (1/(n+k+1))−(1/(n+1))=Σ_(n≥0) ((−k)/((n+1)(n+k+1))) =−k.((Ψ(k+1)−Ψ(1))/k)=Ψ(1)−Ψ(k+1)=−H_k H_k =Σ_(m=1) ^k (1/m) S_m =Σ_(k=0) ^m (((−1)^(k+1) C_m ^k )/(m!))H_k H_k =∫_0 ^1 ((1−x^k )/(1−x))dx Σ(−1)^(k+1) C_n ^k H_k =∫_0 ^1 ΣC_n ^k (−1)^(k+1) ((1−x^k )/(1−x))dx =∫_0 ^1 ((ΣC_n ^k (−x)^k )/(1−x))dx=∫_0 ^1 (((1−x)^m )/(1−x))dx=∫_0 ^1 (1−x)^(m−1) =(1/m) S_m =(1/(m!)).ΣC_m ^k (−1)^(k+1) H_k =(1/(m!)).(1/m)=(1/(m.m!))$
	$S_{m} = \sum_{n ⩾ 1} \frac{1}{n (n + 1) . . (n + k) . . . (n + m)}$ $c o e f i c i e n r o f \frac{1}{(n + k)} i s$ $\frac{{(- 1)}^{k}}{k! (m - k)!} = \frac{1}{m!} . \frac{{(- 1)}^{k} m!}{k! (m - k)!} = \frac{C_{k}^{m} {(- 1)}^{k}}{m!}$ $S_{m} = \sum_{n ⩾ 1} \underset{k = 0}{\sum^{m}} \frac{{(- 1)}^{k} C_{m}^{k}}{m! (n + k)}$ $\underset{k = 0}{\sum^{m}} {(- 1)}^{k} C_{m}^{k} = {(1 - 1)}^{m} = 0$ $S_{m} \Leftrightarrow \sum_{n ⩾ 1} \sum_{k} {{\frac{{(- 1)}^{k} C_{m}^{k}}{m! (n + k)} - \frac{{(- 1)}^{k} C_{m}^{k}}{m! . n}} + \frac{{(- 1)}^{k} C_{m}^{k}}{m! (n + k)}_{= 0}}$ $= \sum_{n ⩾ 1} {\sum_{k ⩾ 0} \frac{{(- 1)}^{k} C_{m}^{k}}{m! (n + k)} - \frac{{(- 1)}^{k} C_{m}^{k}}{m! n}}$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} C_{m}^{k}}{m!} \sum_{n ⩾ 1} \frac{1}{n + k} - \frac{1}{n} . . . . W e c a n c h a n g e \sum_{k} \sum_{n}$ $a b s o l u t C v$ $\sum_{n ⩾ 1} \frac{1}{n + k} - \frac{1}{n} = \sum_{n ⩾ 0} \frac{1}{n + k + 1} - \frac{1}{n + 1} = \sum_{n ⩾ 0} \frac{- k}{(n + 1) (n + k + 1)}$ $= - k . \frac{Ψ (k + 1) - Ψ (1)}{k} = Ψ (1) - Ψ (k + 1) = - H_{k}$ $H_{k} = \underset{m = 1}{\sum^{k}} \frac{1}{m}$ $S_{m} = \underset{k = 0}{\sum^{m}} \frac{{(- 1)}^{k + 1} C_{m}^{k}}{m!} H_{k}$ $H_{k} = \int_{0}^{1} \frac{1 - x^{k}}{1 - x} d x$ $Σ {(- 1)}^{k + 1} C_{n}^{k} H_{k} = \int_{0}^{1} Σ C_{n}^{k} {(- 1)}^{k + 1} \frac{1 - x^{k}}{1 - x} d x$ $= \int_{0}^{1} \frac{Σ C_{n}^{k} {(- x)}^{k}}{1 - x} d x = \int_{0}^{1} \frac{{(1 - x)}^{m}}{1 - x} d x = \int_{0}^{1} {(1 - x)}^{m - 1}$ $= \frac{1}{m}$ $S_{m} = \frac{1}{m!} . Σ C_{m}^{k} {(- 1)}^{k + 1} H_{k} = \frac{1}{m!} . \frac{1}{m} = \frac{1}{m . m!}$
	Commented by SANOGO last updated on 14/Dec/22

	$m e r c i$
	Commented by dre23 last updated on 16/Dec/22

	$j e v o u s e n p r i e$
	Answered by mnjuly1970 last updated on 14/Dec/22
	$S=(1/m)Σ_(n=1) ^∞ {(1/(n(n+1)...(n+m−1)))−(1/((n+1)(n+2)...(n+m)))} = (1/m) ((1/(1.2.3.4...((m−1))))−0) =(1/(m!))$
	$S = \frac{1}{m} \underset{n = 1}{\sum^{\infty}} {\frac{1}{n (n + 1) . . . (n + m - 1)} - \frac{1}{(n + 1) (n + 2) . . . (n + m)}}$ $= \frac{1}{m} (\frac{1}{1.2.3.4 . . . ((m - 1))} - 0)$ $= \frac{1}{m!}$
	Commented by SANOGO last updated on 14/Dec/22

	$m e r c i$
	Commented by dre23 last updated on 15/Dec/22

	$n = 1 w e g e t \frac{1}{m m!}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum