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

Question Number 57847 by Abdo msup. last updated on 13/Apr/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) f i n d {l i m}_{n \to + \infty} u_{n}

Commented by maxmathsup by imad last updated on 14/Apr/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 - 1} \frac{{(- 1)}^{k}}{k^{2}} \Rightarrow

- (u_{1} + u_{2}) + u_{2} + u_{3} - (u_{3} + u_{4}) + \dots . {(- 1)}^{n - 1} (u_{n - 1} + u_{n}) = \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k^{2}} \Rightarrow

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

u_{n} = \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k - n + 1}}{k^{2}} + {(- 1)}^{n - 1} u_{1} w i t h n ⩾ 2

2) t h e s e q u e n c e {(- 1)}^{n} u_{1} i s d i v e r g e n t s o (u_{n}) d i v e r g e s b u t w e c a n s e e

u_{2 n} = - \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} \Rightarrow (u_{2 n}) c o n v e r g e s a l s o

u_{2 n + 1} = \sum_{k = 1}^{2 n} \frac{{(- 1)}^{k}}{k^{2}} + u_{1} \to \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} + u_{1} \Rightarrow u_{2 n + 1} c o n v e r g e s