1-let-U-n-k-0-n-1-k-1-1-1-1-n-1-terms-is-lim-n-U-n-exist-find-U-n-by-using-integr-part-2-let-V-n-k-1-n-k-1-k-1-2-3-4-nterms-is-lim-n-V-n-exist

Question Number 61536 by maxmathsup by imad last updated on 04/Jun/19

1) l e t U_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} = 1 - 1 + 1 - 1 + \dots (n + 1 t e r m s)

i s {l i m}_{n \to + \infty} U_{n} e x i s t ? f i n d U_{n} b y u s i n g i n t e g r p a r t [. .]

2) l e t V_{n} = \sum_{k = 1}^{n} k {(- 1)}^{k} = - 1 + 2 - 3 + 4 + \dots . . (n t e r m s)

i s {l i m}_{n \to + \infty} V_{n} e x i s t

f i n d V_{n} b y u s i n g i n t e g r p a r t [. .]

Commented by maxmathsup by imad last updated on 04/Jun/19

1) w e h a v e f o r x \neq 1 \sum_{k = 0}^{n} x^{k} = \frac{1 - x^{n + 1}}{1 - x} \Rightarrow \sum_{k = 0}^{n} {(- 1)}^{k} = \frac{1 - {(- 1)}^{n + 1}}{2}

\Rightarrow U_{n} = \frac{1}{2} + \frac{{(- 1)}^{n}}{2} t h e s e q u e n c e {(- 1)}^{n} i s n o t c o n v e r g e n t s o U_{n} d o n t c o n v e r g e s

b u t w e h a v e {l i m}_{n \to + \infty} U_{2 n} = 1 a n d {l i m}_{n \to + \infty} U_{2 n + 1} = 0

2) w e h a v e V_{n} = w (- 1) w i t h w (x) = \sum_{k = 1}^{n} k x^{k} = \sum_{k = 0}^{n} {k x}^{k}

w e h a v e \sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} \Rightarrow \sum_{k = 1}^{n} {k x}^{k - 1} = \frac{(n + 1) x^{n} (x - 1) - (x^{n + 1} - 1) \times 1}{{(x - 1)}^{2}}

= \frac{(n + 1) x^{n + 1} - (n + 1) x^{n} - x^{n + 1} + 1}{{(x - 1)}^{2}} = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow

V_{n} = \frac{n {(- 1)}^{n + 1} - (n + 1) {(- 1)}^{n} + 1}{4} = \frac{- n {(- 1)}^{n} - (n + 1) {(- 1)}^{n} + 1}{4}

= \frac{1 - (2 n + 1) {(- 1)}^{n}}{4} i t s c l e a r t h a t V_{n} i s n o t c o n v e r g e n t b u t w e s e e

V_{2 n} = \frac{1 - (2 n + 1)}{4} \to - \infty a n d V_{2 n + 1} = \frac{1 + 2 n + 1}{4} = \frac{n + 1}{2} \to + \infty (n \to + \infty)