U-n-is-a-sequence-wich-verify-U-n-U-n-1-1-n-2-1-find-U-n-interms-of-n-2-calculate-lim-n-U-n-

Question Number 65489 by mathmax by abdo last updated on 30/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} = \frac{1}{n^{2}}

1) f i n d U_{n} i n t e r m s o f n

2) c a l c u l a t e {l i m}_{n \to + \infty} U_{n}

Commented by mathmax by abdo last updated on 03/Aug/19

1) w e h a v e u_{n} + u_{n + 1} = \frac{1}{n^{2}} \Rightarrow \sum_{k = 1}^{n - 1} {(- 1)}^{k} (u_{k} + u_{k + 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}}

\Rightarrow - u_{1} - u_{2} + u_{2} + u_{3} - \dots . + {(- 1)}^{n - 2} (u_{n - 2} + u_{n - 1}) +

{(- 1)}^{n - 1} (u_{n - 1} + u_{n}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} \Rightarrow

- u_{1} + {(- 1)}^{n - 1} u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} \Rightarrow

{(- 1)}^{n - 1} u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} + u_{1} \Rightarrow u_{n} = {(- 1)}^{n - 1} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} + {(- 1)}^{n - 1} u_{1}

2) u_{n} i s n o t c o n v e r g e n t b u t w e s e e t h a t

u_{2 n} = - \sum_{k = 1}^{2 n} \frac{{(- 1)}^{k}}{k^{2}} - u_{1} \to - \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} - u_{1}

u_{2 n + 1} = \sum_{k = 1}^{2 n + 1} \frac{{(- 1)}^{k}}{k^{2}} + u_{1} \to \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} + u_{1}