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-U-n-n-2-

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) 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} + \dots {(- 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 \dots