1-Determine-the-following-if-it-is-convergent-or-divergent-n-1-sin-n-n-2-n-1-sin-n-p-n-p-p-R-find-the-range-of-p-when-it-is-convergent-

Question Number 86113 by Tony Lin last updated on 27/Mar/20

(1) D e t e r m i n e t h e f o l l o w i n g

i f i t i s c o n v e r g e n t o r d i v e r g e n t

\underset{n = 1}{\sum^{\infty}} \frac{s i n (n)}{n}

(2) \underset{n = 1}{\sum^{\infty}} \frac{s i n (n^{p})}{n^{p}}, p ϵ R, f i n d t h e r a n g e

o f p w h e n i t i s c o n v e r g e n t

Commented by Serlea last updated on 27/Mar/20

\underset{n = 1}{\sum^{\infty}} \frac{s i n (n)}{n}

let U_{n} = \frac{\sin (n)}{(n)}

= (\frac{1}{n}) \sin (n)

V_{n} = \frac{n}{n} = 1

\lim_{n \to \infty} \frac{U_{n}}{V_{n}} = \lim_{n \to \infty} \frac{\sin (n)}{n} = 1

Σ U_{n} and Σ V_{n} but converges and diverges

{It}^{'} s diverent from the P - series test

(p = 0 < 1)

2) Σ \frac{\sin (n^{p})}{n^{p}}

U_{n} = \frac{\sin (n^{p})}{n^{p}}

V_{n} = \frac{n^{P}}{n^{p}} = 1

{Lim}_{n \to \infty} \frac{U_{n}}{V_{n}} = \lim_{n \to \infty} \frac{\sin (n^{p})}{n^{p}} let t = n^{p}

= 1

P = 1

Convergent when p ⩾ 1

Commented by mathmax by abdo last updated on 27/Mar/20

1) \sum_{n = 1}^{\infty} \frac{s i n (n)}{n} c o n v e r g e s d u e t o a b e l d i r i c h l e t t h e o r e m

u_{n} = \frac{1}{n} i s d e c e a s i n g t o 0 a n d \exists m > 0 / ∣ \sum_{k = 0}^{n} s i n k ∣< m

Commented by mathmax by abdo last updated on 27/Mar/20

∣ \sum_{n = 1}^{\infty} \frac{s i n (n^{p})}{n^{p}} ∣⩽ \sum_{n = 1}^{\infty} \frac{1}{n^{p}} s o i f p > 1 t h i s s e r i e c o n v e r g e s \dots

Commented by Serlea last updated on 27/Mar/20

I think you are wrong bro

Commented by mathmax by abdo last updated on 27/Mar/20

i a m n o t w r o n g r e v i s e c o u r s e s o f s e r i e s \dots .