a_n is a sequence wich verify a_(n+1) +a_n =(1/(n+1)) ∀n calculate Σ_(n=0) ^∞ a


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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
	$f(x)=Σ_(n=0) ^∞ a_n x^n =Σ_(n=0) ^∞ ((1/(n+1))−a_(n+1) )x^n =Σ_(n=0) ^∞ (x^n /(n+1))−Σ_(n=0) ^∞ a_(n+1) x^n =−((ln(1−x))/x)−(1/x)(Σ_(n=0) ^∞ a_n x^n −a_0 )\ only if ∣x∣<1 =((a_0 −ln(1−x))/x)−((f_n (x))/x) f(x)(1+(1/x))=((a_0 −ln(1−x))/x) f(x)=((a_0 −ln(1−x))/(x+1)) a_0 =1−(1/2)+(1/3)−(1/4)+(1/5)...+_− (1/k)$
	$f (x) = \underset{n = 0}{\sum^{\infty}} a_{n} x^{n} = \underset{n = 0}{\sum^{\infty}} (\frac{1}{n + 1} - a_{n + 1}) x^{n}$ $= \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n + 1} - \underset{n = 0}{\sum^{\infty}} a_{n + 1} x^{n}$ $= - \frac{l n (1 - x)}{x} - \frac{1}{x} (\underset{n = 0}{\sum^{\infty}} a_{n} x^{n} - a_{0}) ∖ o n l y i f ∣ x ∣< 1$ $= \frac{a_{0} - l n (1 - x)}{x} - \frac{f_{n} (x)}{x}$ $f (x) (1 + \frac{1}{x}) = \frac{a_{0} - l n (1 - x)}{x}$ $f (x) = \frac{a_{0} - l n (1 - x)}{x + 1}$ $a_{0} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} . . . \underline{+} \frac{1}{k}$
	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

	$i am fine thank you . . . .$
	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
	$another way we can explicit a_n we have a_n +a_(n+1) =(1/(n+1)) ⇒ Σ_(k=0) ^n (−1)^k (a_k +a_(k+1) ) =Σ_(k=0) ^n (((−1)^k )/(k+1)) ⇒ (a_0 +a_1 )−(a_1 +a_2 )+.....(−1)^(n−1) (a_(n−1) +a_n ) +(−1)^n (a_(n ) +a_(n+1) ) =Σ_(k=0) ^n (((−1)^k )/(k+1)) ⇒a_0 +(−1)^n a_(n+1) =Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) (−1)^n a_(n+1) =−a_0 −Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) ⇒ a_(n+1) =(−1)^(n+1) a_0 +(−1)^(n+1) Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) ⇒ a_n =(−1)^n a_0 +(−1)^n Σ_(k=1) ^n (((−1)^(k−1) )/k) (for n>0) ⇒ Σ_(n=0) ^∞ a_n x^n =a_0 Σ_(n=0) ^∞ (−1)^n x^n +Σ_(n=0) ^∞ (−1)^n {Σ_(k=1) ^(n ) (((−1)^(k−1) )/k)} x^n =(a_0 /(x+1)) +Σ_(n=0) ^∞ (−1)^(n ) { Σ_(k=1) ^n (((−1)^(k−1) )/k)}x^n$
	$another way we can explicit a_{n} we have a_{n} + a_{n + 1} = \frac{1}{n + 1} \Rightarrow$ $\sum_{k = 0}^{n} {(- 1)}^{k} (a_{k} + a_{k + 1}) = \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} \Rightarrow$ $(a_{0} + a_{1}) - (a_{1} + a_{2}) + . . . . . {(- 1)}^{n - 1} (a_{n - 1} + a_{n}) + {(- 1)}^{n} (a_{n} + a_{n + 1})$ $= \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} \Rightarrow a_{0} + {(- 1)}^{n} a_{n + 1} = \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k}$ ${(- 1)}^{n} a_{n + 1} = - a_{0} - \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} \Rightarrow$ $a_{n + 1} = {(- 1)}^{n + 1} a_{0} + {(- 1)}^{n + 1} \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} \Rightarrow$ $a_{n} = {(- 1)}^{n} a_{0} + {(- 1)}^{n} \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} (for n > 0) \Rightarrow$ $\sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} + \sum_{n = 0}^{\infty} {(- 1)}^{n} {\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k}} x^{n}$ $= \frac{a_{0}}{x + 1} + \sum_{n = 0}^{\infty} {(- 1)}^{n} {\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k}} x^{n}$
	Commented by mathmax by abdo last updated on 05/Jun/20

	$∣ x ∣< 1$
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