U_n is a sequence wich verify U_n +U_(n+1) =n for all integr n 1) calculate U_n intrem of n 2) find nature of the serie Σ (U


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Question Number 65193 by mathmax by abdo last updated on 26/Jul/19

$U_{n} i s a s e q u e n c e w i c h v e r i f y U_{n} + U_{n + 1} = n f o r a l l i n t e g r n$ $1) c a l c u l a t e U_{n} i n t r e m o f n$ $2) f i n d n a t u r e o f t h e s e r i e Σ \frac{U_{n}}{n^{2}}$
Commented by mathmax by abdo last updated on 26/Jul/19
$1) we have U_n +U_(n+1) =n ⇒Σ_(k=0) ^(n−1) (−1)^k (U_k +U_(k+1) ) =Σ_(k=0) ^(n−1) k(−1)^k ⇒U_0 +U_1 −U_1 −U_2 +...(−1)^(n−2) (U_(n−2) +U_(n−1) )+(−1)^(n−1) (U_(n−1) +U_n ) =Σ_(k=0) ^(n−1) k(−1)^k ⇒ U_0 +(−1)^(n−1) U_n =Σ_(k=0) ^(n−1) k(−1)^k ⇒(−1)^(n−1) U_n =Σ_(k=0) ^(n−1) k(−1)^k −U_0 ⇒U_n =Σ_(k=0) ^(n−1) k(−1)^(k+n−1) −(−1)^(n−1) U_0 ⇒ U_n =(−1)^(n−1) Σ_(k=0) ^(n−1) k(−1)^k +(−1)^n U_0 let p(x) =Σ_(k=0) ^(n−1) kx^k we have Σ_(k=0) ^(n−1) x^k =((1−x^n )/(1−x)) (x≠1) ⇒ Σ_(k=1) ^(n−1) kx^(k−1) =(((x^n −1)/(x−1)))^′ =((nx^(n−1) (x−1)−(x^n −1)×1)/((x−1)^2 )) =((nx^n −nx^(n−1) −x^n +1)/((x−1)^2 )) =(((n−1)x^n −nx^(n−1) +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^(n−1) kx^k =(((n−1)x^(n+1) −nx^n +x)/((x−1)^2 )) ⇒ Σ_(k=0) ^(n−1) k(−1)^k =(((n−1)(−1)^(n+1) −n(−1)^n −1)/4) =((−n(−1)^n +(−1)^n −n(−1)^n −1)/4) =((−2n(−1)^n +(−1)^n −1)/4) ⇒ U_n =(((−1)^(n−1) )/4){ −2n(−1)^n +(−1)^n −1}+(−1)^n U_0 =−(1/4){−2n+1−(−1)^n }+(−1)^n U_0 ⇒ U_n =(1/4){2n−1 +(−1)^n } +(−1)^n U_0 2) Σ_(n=1) ^∞ (U_n /n^2 ) =Σ_(n=1) ^∞ (1/(2n)) −(1/4) Σ_(n=1) ^∞ (1/n^2 ) +(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +Σ_(n=1) ^∞ (((−1)^n U_0 )/n^2 ) the serie Σ(1/(2n)) diverges ⇒Σ(U_n /n^2 ) diverges...$
$1) w e h a v e U_{n} + U_{n + 1} = n \Rightarrow \sum_{k = 0}^{n - 1} {(- 1)}^{k} (U_{k} + U_{k + 1}) = \sum_{k = 0}^{n - 1} k {(- 1)}^{k}$ $\Rightarrow U_{0} + U_{1} - U_{1} - U_{2} + . . . {(- 1)}^{n - 2} (U_{n - 2} + U_{n - 1}) + {(- 1)}^{n - 1} (U_{n - 1} + U_{n})$ $= \sum_{k = 0}^{n - 1} k {(- 1)}^{k} \Rightarrow$ $U_{0} + {(- 1)}^{n - 1} U_{n} = \sum_{k = 0}^{n - 1} k {(- 1)}^{k} \Rightarrow {(- 1)}^{n - 1} U_{n} = \sum_{k = 0}^{n - 1} k {(- 1)}^{k} - U_{0}$ $\Rightarrow U_{n} = \sum_{k = 0}^{n - 1} k {(- 1)}^{k + n - 1} - {(- 1)}^{n - 1} U_{0} \Rightarrow$ $U_{n} = {(- 1)}^{n - 1} \sum_{k = 0}^{n - 1} k {(- 1)}^{k} + {(- 1)}^{n} U_{0}$ $l e t p (x) = \sum_{k = 0}^{n - 1} {k x}^{k} w e h a v e \sum_{k = 0}^{n - 1} x^{k} = \frac{1 - x^{n}}{1 - x} (x \neq 1) \Rightarrow$ $\sum_{k = 1}^{n - 1} {k x}^{k - 1} = {(\frac{x^{n} - 1}{x - 1})}^{^{'}} = \frac{{n x}^{n - 1} (x - 1) - (x^{n} - 1) \times 1}{{(x - 1)}^{2}}$ $= \frac{{n x}^{n} - {n x}^{n - 1} - x^{n} + 1}{{(x - 1)}^{2}} = \frac{(n - 1) x^{n} - {n x}^{n - 1} + 1}{{(x - 1)}^{2}} \Rightarrow$ $\sum_{k = 1}^{n - 1} {k x}^{k} = \frac{(n - 1) x^{n + 1} - {n x}^{n} + x}{{(x - 1)}^{2}} \Rightarrow$ $\sum_{k = 0}^{n - 1} k {(- 1)}^{k} = \frac{(n - 1) {(- 1)}^{n + 1} - n {(- 1)}^{n} - 1}{4}$ $= \frac{- n {(- 1)}^{n} + {(- 1)}^{n} - n {(- 1)}^{n} - 1}{4} = \frac{- 2 n {(- 1)}^{n} + {(- 1)}^{n} - 1}{4} \Rightarrow$ $U_{n} = \frac{{(- 1)}^{n - 1}}{4} {- 2 n {(- 1)}^{n} + {(- 1)}^{n} - 1} + {(- 1)}^{n} U_{0}$ $= - \frac{1}{4} {- 2 n + 1 - {(- 1)}^{n}} + {(- 1)}^{n} U_{0} \Rightarrow$ $U_{n} = \frac{1}{4} {2 n - 1 + {(- 1)}^{n}} + {(- 1)}^{n} U_{0}$ $2) \sum_{n = 1}^{\infty} \frac{U_{n}}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{2 n} - \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} U_{0}}{n^{2}}$ $t h e s e r i e Σ \frac{1}{2 n} d i v e r g e s \Rightarrow Σ \frac{U_{n}}{n^{2}} d i v e r g e s . . .$
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