Is-the-following-series-convergent-i-1-sin-i-i-

Question Number 2722 by prakash jain last updated on 25/Nov/15

Is the following series convergent

\underset{i = 1}{\sum^{\infty}} \frac{\sin i}{i}

Answered by Filup last updated on 25/Nov/15

no

for x \in (- \infty, \infty)

- 1 ⩽ s i n x ⩽ 1

Answered by prakash jain last updated on 25/Nov/15

\int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}

\int_{1}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2} - Si (1)

As the integral converges, the series is

convergent .

Update : Wrong test used . The above

argument does not prove covergence .

Commented by Yozzi last updated on 26/Nov/15

I c o n j e c t u r e d t h a t i f w e l e t

u_{n} = \frac{s i n i}{i} + \frac{s i n (i + 1)}{i + 1} + \frac{s i n (i + 2)}{i + 2} a n d

u_{n + 1} = \frac{s i n (i + 3)}{i + 3} + \frac{s i n (i + 4)}{i + 4} + \frac{s i n (i + 5)}{i + 5}

f o r \underset{i = 1}{\sum^{\infty}} \frac{s i n i}{i} = u_{1} + u_{2} + u_{3} + \dots

{u_{n}} i s a n a l t e r n a t i n g s e q u e n c e .

Y e s . I t h i n k y o u r u_{j} i s a l t e r n a t i n g .

Commented by prakash jain last updated on 26/Nov/15

\underset{n = 1}{\sum^{\infty}} \frac{\sin n}{n} = \underset{j = 1}{\sum^{\infty}} u_{j}

u_{j} = \underset{k = ⌈ π (j - 1) ⌉}{\sum^{⌊ π j ⌋}} \frac{\sin k}{k}

is \underset{j = 1}{\sum^{\infty}} u_{j} an alternating series ?

Commented by Filup last updated on 25/Nov/15

M y m i s t a k e . I Was thinking

incorrectly

i was thinking of Σ (\sin x)

Commented by Yozzi last updated on 25/Nov/15

f (x) = \frac{s i n x}{x} i s n o t a p o s i t i v e, m o n o t o n i c a l l y

d e c r e a s i n g f u n c t i o n \forall x \in [1, \infty) t o

u s e t h e i n t e g r a l t e s t .

Commented by prakash jain last updated on 25/Nov/15

Ok . You are right .

How to prove that \underset{i = 1}{\sum^{\infty}} \sin (x) / x is convergent ?

Commented by Yozzi last updated on 25/Nov/15

A l t e r n a t i n g s e r i e s t e s t :

F o r a n i n f i n i t e s e r i e s \underset{n = 1}{\sum^{\infty}} u_{n} o f a l t e r n a t i n g

t e r m s t h e f o l l o w i n g c o n d i t i o n s a r e

r e q u i r e d f o r c o n v e r g e n c e :

(1) ∣ u_{n + 1} ∣⩽∣ u_{n} ∣ f o r n ⩾ N

(2) \lim_{n \to \infty} u_{n} = 0 (o r \lim_{n \to \infty} ∣ u_{n} ∣= 0)

Commented by prakash jain last updated on 25/Nov/15

The series is not alternating either .

Commented by Yozzi last updated on 25/Nov/15

t r u e \dots

Commented by prakash jain last updated on 26/Nov/15

Based on your suggestion I was trying to

come up with an equivalent alternating

series .

Answered by Yozzi last updated on 26/Nov/15

C o n s i d e r t h e s e r i e s T = \underset{i = 1}{\sum^{\infty}} \frac{s i n i x}{i} (x \neq \pm π) .

{D i r i c h l e t}^{'} s t e s t m a y b e u s e d t o p r o v e

t h a t T i s u n i f o r m l y c o n v e r g e n t i n

a n i n t e r v a l a ⩽ x ⩽ b .

S u p p o s e t h a t :

(1) {a_{n}} i s a s e q u e n c e o f p o s i t i v e

c o n s t a n t s t h a t a r e m o n o t o n i c

d e c r e a s i n g, h a v i n g a l i m i t o f z e r o;

(2) \exists P (c o n s t a n t) s u c h t h a t f o r

a ⩽ x ⩽ b, ∣ \underset{r = 1}{\sum^{n}} u_{r} (x) ∣< P, \forall n > N .

T h e n, t h e s e r i e s \underset{n = 1}{\sum^{\infty}} a_{n} u_{n} (x) i s

u n i f o r m l y c o n v e r g e n t o n t h e i n t e r v a l

a ⩽ x ⩽ b .

L e t u_{i} (x) = s i n i x, a_{i} = \frac{1}{i} (i \in N) .

(1) F o r t h e s e q u e n c e {a_{i}}, i t s t e r m s

a r e m o n o t o n i c d e c r e a s i n g a n d

p o s i t i v e .

\lim_{i \to \infty} a_{i} = \lim_{i \to \infty} \frac{1}{i} = \frac{1}{\infty} = 0 .

(2) S u p p o s e 1 ⩽ x ⩽ \frac{π}{2} a n d d e f i n e S = \underset{i = 1}{\sum^{n}} u_{i} (x) .

(s i n \frac{x}{2}) S = \underset{r = 1}{\sum^{n}} s i n i x s i n \frac{x}{2} . (1 ⩽ x ⩽ \frac{π}{2})

S i n c e - 2 s i n i x s i n \frac{x}{2} = c o s (i + \frac{1}{2}) x - c o s (i - \frac{1}{2}) x

\Rightarrow s i n i x s i n \frac{x}{2} = \frac{- 1}{2} {f (r) - f (r - 1)}

w h e r e f (r) = c o s (i + \frac{1}{2}) x (i \in N) .

∴ S s i n \frac{x}{2} = \frac{- 1}{2} \underset{i = 1}{\sum^{n}} {f (r) - f (r - 1)}

- 2 S s i n \frac{x}{2} = f (1) - f (0) + f (2) - f (1) +

f (3) - f (2) + f (4) - f (3) + f (5) - f (4) +

\dots + f (n - 3) - f (n - 4) + f (n - 2) - f (n - 3)

+ f (n - 1) - f (n - 2) + f (n) - f (n - 1)

- 2 S s i n \frac{x}{2} = f (n) - f (0)

\Rightarrow S = \frac{f (0) - f (n)}{2 s i n \frac{x}{2}}

S = \frac{c o s \frac{x}{2} - c o s (n + \frac{1}{2}) x}{2 s i n \frac{x}{2}}

- (c o s (n + \frac{1}{2}) x - c o s \frac{x}{2})

= - (- 2 s i n \frac{n + \frac{1}{2} + \frac{1}{2}}{2} x s i n \frac{n + \frac{1}{2} - \frac{1}{2}}{2} x)

= 2 s i n \frac{n + 1}{2} x s i n \frac{n x}{2}

∴ S = \frac{s i n \frac{n x}{2} s i n \frac{n + 1}{2} x}{s i n \frac{x}{2}} = (\frac{1}{s i n \frac{x}{2}}) s i n \frac{n x}{2} s i n \frac{n + 1}{2} x

∴∣ S ∣=∣ \frac{1}{s i n \frac{x}{2}} s i n \frac{n + 1}{2} s i n \frac{n x}{2} ∣

S i n c e 1 ⩽ x ⩽ \frac{π}{2} \Rightarrow \frac{1}{2} ⩽ \frac{x}{2} ⩽ \frac{π}{4} \Rightarrow s i n \frac{x}{2} ⩾ s i n \frac{1}{2} f o r 1 ⩽ x ⩽ \frac{π}{2} .

∴∣ S ∣= c o s e c \frac{x}{2} ∣ s i n \frac{n x}{2} s i n \frac{n + 1}{2} x ∣

N o w ∣ s i n \frac{u}{2} ∣< 1 f o r \forall u \in R

\Rightarrow c o s e c \frac{x}{2} ∣ s i n \frac{n x}{2} s i n \frac{n + 1}{2} x ∣< \frac{1}{s i n \frac{x}{2}} \forall n > 1

\Rightarrow ∣ S ∣< \frac{1}{s i n \frac{x}{2}} \forall n > 1 (N = 1)

m a x (s i n \frac{x}{2}) f o r 1 ⩽ x ⩽ \frac{π}{2} i s s i n \frac{π}{4} = \frac{1}{\sqrt{2}}

\Rightarrow∣ S ∣< \sqrt{2} \Rightarrow P = \sqrt{2} f o r 1 ⩽ x ⩽ \frac{π}{2} .

S o, c o n d i t i o n s (1) a n d (2) a r e s a t i s f i e d .

H e n c e, b y {D i r i c h l e t}^{'} s t e s t,

\underset{i = 1}{\sum^{\infty}} \frac{s i n i x}{i} i s u n i f o r m l y c o n v e r g e n t

i n t h e i n t e r v a l 1 ⩽ x ⩽ \frac{π}{2} . H e n c e, i f x = 1,

\underset{i = 1}{\sum^{\infty}} \frac{s i n i}{i} i s u n i f o r m l y c o n v e r g e n t .

(I t r i e d p a r t (2) b u t I k n o w t h e a p p r o a c h

i s i n c o r r e c t .)

Commented by prakash jain last updated on 26/Nov/15

Innovative!

The series in question could start

with \sin i \sin 0.5 (x = 1) .

However you have proved more generic case .

Your approach looks correct to me .

Why do you think it is wrong ?

Commented by Yozzi last updated on 26/Nov/15

T h i s a r e a i s n e w t o m e a n d I {h a v e n}^{'} t

m a s t e r e d m e t h o d o f p r o o f y e t . B u t

I^{'} m g l a d y o u s e e n o i s s u e s w i t h i t :) .

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