Sin( http://math.stackexchange.com/questions/238997

[Pages:3]Prove the divergence of the sequence $\left\{ \sin(...

...

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Prove the divergence of the sequence {sin(n)} n=1.

I am looking for nice ways of proving the divergence of the sequence {xn} n=1 defined by xn = sin (n).

One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals

Ik

=

(

6

+ 2(k -

1),

5 6

+ 2(k - 1))

and one where they lie in

Jk

=

(

7 6

+ 2(k - 1),

11 6

+ 2(k -

1)) .

If xn converges, then all has its values in [-1, -

its subsequences must

1 2

].

Contradiction.

converge

to

the

same

limit,

but

here

the

first

subsequence

has

all

its

values

in

the

interval

[

1 2

,

1]

while

the

second

Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?

( sequences-and-series )

edited Oct 21 at 14:33 hlapointe 365 1 12

asked Nov 17 '12 at 1:54 Spenser 7,850 2 15 49

Hi, Sos440 Is it possible to extend this method for a proof of divergence of sin(n!)? ? M.R. Yegan Oct 1

'14 at 4:43

7 Answers

Equidistribution argument is very elegant, but it becomes a sledgehammer method when it comes to a question of mere convergence.

A simple argument can reveal the divergence of (sin n, n 1). Let

(xn, yn) = (cos n, sin n)

(I changed the notation for the sake of consistency of notation.) Then the application of the addition formula for trigonometric function or the rotation matrix gives

xn+1 = xn cos 1 - yn sin 1 yn+1 = xn sin 1 + yn cos 1. Now assume (yn) converges. Then since sin 1 0, we have

xn+1 = (yn+1 - yn cos 1) cot 1 - yn sin 1 and hence (xn) also converges. Now let (xn, yn) (, ). Then taking limit to the recursive

formula we have

= cos 1 - sin 1 = sin 1 + cos 1. Solving this system of linear equations give (, ) = (0, 0). On the other hand, since

x2n + yn2 = 1,

we must have

2 + 2 = 1, a contradiction! Therefore (yn) cannot converge. ////

Of course, we can say much more on (yn). For example, we can show that the set of limit points of (yn) is exactly [-1, 1], and the Cesaro mean of (yn) is 0 from Weyl's criterion.

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Prove the divergence of the sequence $\left\{ \sin(...

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(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)

edited Nov 17 '12 at 2:50

answered Nov 17 '12 at 2:30 Sangchul Lee 44.5k 4 94 155

This is very nice! ? Jason DeVito Nov 17 '12 at 2:42

1 Very nice indeed! Thanks! ? Spenser Nov 17 '12 at 2:45

Thanks! Anyway I forgot to comment that this is not my original idea, but the idea that I learned from the book Problems in Real Analysis Advanced Calculus on the Real Axis. ? Sangchul Lee Nov 17 '12 at 2:49

Very nice argument! So this also proves that (xn) = (cos n) cannot converge. ? Paul Apr 16 '13 at 21:21

1

Another quick way to see that, if

to note that n=0 sin(n) and

lim nn=0 cossi(nn()nh)aavendboliumndnedpacrtoiasl(snum) es.xiIsntd, etheedn,

both limits are zero is

nk=0 sin(k) and

nk=0 cos(k) are the real and imaginary parts of nk=0 eik

=

ei(n+1) -1 ei -1

which is bounded. ? Mike F

May 26 '14 at 8:34

If is an irrational number, the numbers n, considered mod 1, are dense in [0, 1]. (A stronger result is that they are equidistributed.) Thus the numbers 2n/ are dense in [0, 1], and therefore the integers are dense mod /2. It follows that sin n diverges.

answered Nov 17 '12 at 2:03 Bruno Joyal 34.6k 5 61 137

This question was posed today (April 16, 2013) as and Douglas Zare gave a nice trig-identity-based proof in comments there.

answered Apr 16 '13 at 21:13 Barry Cipra 11 1

Essentially, every point in the interval [-1, 1] is a limit point for the sequence {sin n}. Since there is

more than one limit point, the sequence diverges.

answered Nov 17 '12 at 2:19 glebovg 5,106 2 14 35

Definition: [x] is a rounding function if [x] is the biggest integer such that for a real number x, x Theorem 1: Define x_n=[10^n x]/10^n for a real number x. Then lim(n)x_n=x. Proof: |x_n-x|=|([10^n x]-10^n x)/10^n |1/10^n Theorem 2: |[10^n ]-10^n | diverges. Proof: Suppose |[10^n ]-10^n | converges to a real number. This implies that there are integers k between 0 and 10 and K such that for nK, the nth decimal place of is k. This is absurd since is irrational. Theorem 3: sin(n) diverges. Proof: If sin[10^n x] has a limit s, then for any >0 there exist an integer K such that for nK, |ssin[10^n x]|< . However, this implies that |[10^n ]-10^n | converges which is proven to

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Prove the divergence of the sequence $\left\{ \sin(...

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be false.

answered Apr 24 '13 at 15:54 Applicable Nho 33 3

Imagine marching around the circumference of the unit circle in steps of arc length 1. Then sin(n) is

the y coordinate at the nth step. Since > 1, the y coordinate will be positive infinitely often and

negative infinitely often. Consequently the limit of sin(n), if it exists, would have to be 0. But for the

same limit,

riefaist oenxi(sts,>wo1u)l,dsihna(vne )tobesaitnl(eas-2t 1si)n=( s-2i1n)(,w-hich-2i1s

) for infinitely greater than 0.

many n, and hence the

These contradictory

requirements show that the limit does not exist.

Remark: This proof does not rely on being an irrational number; nor does it use any trig identities

beyond the simplest symmetries of the sine function.

edited Oct 21 at 2:46

answered Oct 20 at 23:29 Barry Cipra 25.2k 1 19 61

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