existant-and-value-m-1-of-S-m-n-1-oo-1-n-n-1-n-m-

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) \dots (n + m)}

Answered by dre23 last updated on 15/Dec/22

S_{m} = \sum_{n ⩾ 1} \frac{1}{n (n + 1) . . (n + k) \dots (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} \dots . 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 = \frac{1}{m} \underset{n = 1}{\sum^{\infty}} {\frac{1}{n (n + 1) \dots (n + m - 1)} - \frac{1}{(n + 1) (n + 2) \dots (n + m)}}

= \frac{1}{m} (\frac{1}{1.2.3.4 \dots ((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!}

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