Calculate-lim-n-A-n-0-1-x-n-1-x-dx-

Question Number 172307 by mathocean1 last updated on 25/Jun/22

C a l c u l a t e

Double subscripts: use braces to clarify

Answered by aleks041103 last updated on 25/Jun/22

A_{n} = \int_{0}^{1} \frac{x^{n}}{1 - (- x)} d x = {(- 1)}^{n} \int_{0}^{1} \frac{{(- x)}^{n} - 1 + 1}{1 - (- x)} d x =

= {(- 1)}^{n + 1} \int_{0}^{1} \frac{1 - {(- x)}^{n} - 1}{1 - (- x)} d x =

= {(- 1)}^{n + 1} [\int_{0}^{1} \frac{1 - {(- x)}^{n}}{1 - (- x)} d x - \int_{0}^{1} \frac{d x}{1 + x}] =

= {(- 1)}^{n + 1} [\int_{0}^{1} (\underset{k = 1}{\sum^{n}} {(- x)}^{k - 1}) d x - l n (1 + 1) + l n (1 + 0)] =

= {(- 1)}^{n + 1} [\underset{k = 1}{\sum^{n}} {(- 1)}^{k - 1} (\int_{0}^{1} x^{k - 1} d x) - l n (2)] =

= {(- 1)}^{n + 1} [\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k - 1}}{k} - l n (2)]

\frac{1}{1 + x} = \underset{k = 0}{\sum^{\infty}} {(- x)}^{k}

\Rightarrow \int_{0}^{s} \frac{d x}{1 + x} = l n (1 + s) = \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \frac{s^{k + 1}}{k + 1} =

\Rightarrow l n (1 + s) = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1} s^{k}}{k}

\Rightarrow \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k - 1}}{k} = l n (2) - \underset{k = n + 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k}

\Rightarrow A_{n} = {(- 1)}^{n} \underset{k = n + 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k}

Double subscripts: use braces to clarify

Answered by Mathspace last updated on 25/Jun/22

A_{n} = \int_{R^{+}} \frac{x^{n}}{1 + x} χ_{[0, 1]} (x) d x

= \int_{R^{+}} f_{n} (x) d x

f_{m} c o n v e r g e s i m p l e m e n t v e r s 0

c a r x \in [0, 1] a n d f_{n} e s t d o m i n e e p a r

g (x) = \frac{1}{1 + x} \Rightarrow l i m A_{n} = \int_{R^{+}} l i m f_{n} = 0

Answered by puissant last updated on 25/Jun/22

0 ⩽ x ⩽ 1 \Rightarrow 1 ⩽ 1 + x ⩽ 2

o n a : \frac{x^{n}}{2} ⩽ \frac{x^{n}}{1 + x} ⩽ x^{n}

\Rightarrow \frac{1}{2} \int_{0}^{1} x^{n} d x ⩽ I_{n} ⩽ \int_{0}^{1} x^{n} d x

\Rightarrow \frac{1}{2 (n + 1)} ⩽ I_{n} ⩽ \frac{1}{n + 1}

d o n c \lim_{n \to + \infty} I_{n} = 0

d^{'} a p r e s l e t h e o r e m e d e s g e n d a r m e s . .

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