a-n-is-a-sequence-wich-verify-a-n-1-a-n-1-n-1-n-calculate-n-0-a-n-x-n-

Question Number 96836 by mathmax by abdo last updated on 05/Jun/20

a_{n} is a sequence wich verify a_{n + 1} + a_{n} = \frac{1}{n + 1} \forall n

calculate \sum_{n = 0}^{\infty} a_{n} x^{n}

Answered by Smail last updated on 05/Jun/20

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

= \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n + 1} - \underset{n = 0}{\sum^{\infty}} a_{n + 1} x^{n}

= - \frac{l n (1 - x)}{x} - \frac{1}{x} (\underset{n = 0}{\sum^{\infty}} a_{n} x^{n} - a_{0}) ∖ o n l y i f ∣ x ∣< 1

= \frac{a_{0} - l n (1 - x)}{x} - \frac{f_{n} (x)}{x}

f (x) (1 + \frac{1}{x}) = \frac{a_{0} - l n (1 - x)}{x}

f (x) = \frac{a_{0} - l n (1 - x)}{x + 1}

a_{0} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} \dots \underline{+} \frac{1}{k}

Commented by mathmax by abdo last updated on 05/Jun/20

how are you sir smail you are absent for a long time in this forum

you still stand in usa ?

Commented by Smail last updated on 05/Jun/20

Y e s, I a m s t i l l i n t h e U S A .

I b o u g h t a n e w p h o n e a n d I {d i d n}^{'} t a c h a n c e

t o i n s t a l l t h i s a p p u n t i l t w o w e e k s a g o .

Commented by Smail last updated on 05/Jun/20

H o w a b o u t y o u ? H o w i s y o u r l i f e w i t h t h i s

c o r o n a c r i s i s ?

Commented by mathmax by abdo last updated on 05/Jun/20

i am fine thank you \dots .

Answered by mathmax by abdo last updated on 05/Jun/20

we have a_{n + 1} + a_{n} = \frac{1}{n + 1} \Rightarrow a_{n + 1} = \frac{1}{n + 1} - a_{n} \Rightarrow \sum_{n = 0}^{\infty} a_{n + 1} x^{n}

= \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 1} - \sum_{n = 0}^{\infty} a_{n} x^{n} = \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} - \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow

\sum_{n = 1}^{\infty} a_{n} x^{n - 1} = \frac{1}{x} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow

\frac{1}{x} \sum_{n = 1}^{\infty} a_{n} x^{n} + \sum_{n = 0}^{\infty} a_{n} x^{n} = - \frac{1}{x} \ln (1 - x) \Rightarrow

\frac{1}{x} (\sum_{n = 0}^{\infty} a_{n} x^{n} - a_{0}) + \sum_{n = 0}^{\infty} a_{n} x^{n} = - \frac{\ln (1 - x)}{x} \Rightarrow

(\frac{1}{x} + 1) \sum_{n = 0} a_{n} x^{n} = \frac{a_{0}}{x} - \frac{\ln (1 - x)}{x} \Rightarrow (x + 1) \sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} - \ln (1 - x) \Rightarrow

\sum_{n = 0}^{\infty} a_{n} x^{n} = \frac{a_{0}}{x + 1} - \frac{\ln (1 - x)}{x + 1}

Commented by mathmax by abdo last updated on 05/Jun/20

another way we can explicit a_{n} we have a_{n} + a_{n + 1} = \frac{1}{n + 1} \Rightarrow

\sum_{k = 0}^{n} {(- 1)}^{k} (a_{k} + a_{k + 1}) = \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} \Rightarrow

(a_{0} + a_{1}) - (a_{1} + a_{2}) + \dots . . {(- 1)}^{n - 1} (a_{n - 1} + a_{n}) + {(- 1)}^{n} (a_{n} + a_{n + 1})

= \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} \Rightarrow a_{0} + {(- 1)}^{n} a_{n + 1} = \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k}

{(- 1)}^{n} a_{n + 1} = - a_{0} - \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} \Rightarrow

a_{n + 1} = {(- 1)}^{n + 1} a_{0} + {(- 1)}^{n + 1} \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} \Rightarrow

a_{n} = {(- 1)}^{n} a_{0} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} (for n > 0) \Rightarrow

\sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} + \sum_{n = 0}^{\infty} {(- 1)}^{n} {\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k}} x^{n}

= \frac{a_{0}}{x + 1} + \sum_{n = 0}^{\infty} {(- 1)}^{n} {\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k}} x^{n}

Commented by mathmax by abdo last updated on 05/Jun/20

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