U-n-is-a-sequence-wich-verify-n-N-U-n-U-n-1-1-n-1-calculate-U-n-interms-of-n-2-is-the-sequence-U-n-convergent-

Question Number 65488 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 \forall n \in N^{★}

U_{n} + U_{n + 1} = \frac{1}{n}

1) c a l c u l a t e U_{n} i n t e r m s o f n

2) i s t h e s e q u e n c e U_{n} c o n v e r g e n t ?

Commented by mathmax by abdo last updated on 31/Jul/19

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

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

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

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

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

2) t h e s e q u e n c e {(- 1)}^{n - 1} i s n o t c o n v e r g e n t e s o 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 - 1} \frac{{(- 1)}^{k}}{k} - u_{1} \to l n (2) - u_{1}

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