prove-that-cos-n-sin-n-n-converge-sequence-

Question Number 129997 by mohammad17 last updated on 21/Jan/21

p r o v e t h a t [(c o s (n) + s i n (n)) / n] c o n v e r g e s e q u e n c e

Answered by Dwaipayan Shikari last updated on 21/Jan/21

\underset{n = 1}{\sum^{\infty}} \frac{c o s n + s i n n}{n} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{e^{i n}}{n} + \frac{e^{- i n}}{n} + \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{e^{i n}}{n} - \frac{e^{- i n}}{n}

= - \frac{1}{2} (l o g (1 - 2 c o s 1 + 1) + l o g (\frac{1 - e^{i}}{1 - e^{- i}})

= \frac{1}{2} l o g (\frac{1}{4 {s i n}^{2} \frac{1}{2}}) + \frac{π}{2} - \frac{1}{2} = \frac{1}{2} (π - 1 - l o g (4 {s i n}^{2} \frac{1}{2})) \sim 1.11

Commented by mohammad17 last updated on 21/Jan/21

t h a n k y o u s i r b u t i w a n t t h i s b y a n a l y e t i c f u n c t i o n

Answered by mathmax by abdo last updated on 21/Jan/21

let u_{n} = \frac{\cos (n) + \sin (n)}{n} \Rightarrow u_{n} = \frac{\sqrt{2} \cos (n - \frac{π}{4})}{n}

\Rightarrow \sum_{n = 1}^{\infty} u_{n} = \sqrt{2} \sum_{n = 1}^{\infty} \frac{\cos (n - \frac{π}{4})}{n} = \sqrt{2} \sum_{n = 1}^{\infty} α_{n} . β_{n}

α_{n} decrease to and we can determine m > 0 /

∣ \sum_{k = 1}^{n} β_{k} ∣⩽ m so the convergence is assured by abel dirichlet

theorem . .

Commented by mathmax by abdo last updated on 21/Jan/21

α_{n} decrease to 0

Commented by mohammad17 last updated on 21/Jan/21

b u t s i r t h i s i s s e q u e n c e n o t s e r i e s a n d m y

t e a c h e r s a y d m e i t w a n t t h e s o l u t i o n b y

\forall \in> 0 \exists k \in N S u c h t h a t ∣ x_{n} - x_{0} ∣<\in \forall n > k

c a n h e l p m e ? s i r

Commented by mathmax by abdo last updated on 22/Jan/21

your teacher want to turn you to stone era \dots no need to use ξ

look we have \frac{cosn + sinn}{n} = \frac{\sqrt{2} \cos (n - \frac{π}{4})}{n} \Rightarrow

∣ \frac{\cos (n) + \sin (n)}{n} ∣⩽ \frac{\sqrt{2}}{n} \to 0 \Rightarrow \lim_{n \to + \infty} \frac{cosn + sinn}{n} = 0