Does the Series \(\sum_n (\sin(n))^n\) Converge?

  • Thread starter Take_it_Easy
  • Start date
In summary: I'd say it was just an inquisitive student :)In summary, the conversation discusses the convergence of the series \sum_n (\sin(n))^n and different approaches to proving whether it converges or not. One approach involves looking at the limit of the terms as n goes to infinity, while another involves considering the absolute value of the terms. The conversation also briefly touches on the topic of approximating pi by rationals and its connection to the convergence of the series. Ultimately, the question of whether the series converges or not remains unresolved.
  • #1
Take_it_Easy
41
0
This is not an homework, but just a couriosity that I have
I never found a easy way to solve the question of convergence of this serie

[tex]\sum_n (\sin(n))^n[/tex]

any help is appreciated!
 
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  • #2
look at exp(in)=cos(n)+isin(n)
(exp(in))^n=exp(in^2)=cos(n^2)+isin(n^2)=(cos(n)+isin(n))^n
Now because exp(in^2) converges also sin(n)^n converges.

I hope I haven't done a mess with the question here.
 
  • #3
Based on a maple plot I have made, it appears not to converge.
Plotted is
[tex]f(X) = \sum_{n=1}^{n=X}\sin{n}^n[/tex]
from X=0,100000
 

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  • #4
Or, i guess we could have also reasoned this way:

SInce |sin(n)|<1 for all n positive integers. then the series

[tex]\sum_{n=1}^{\infty}|sin(n)|^n [/tex] converges, so the original series converges absolutely=> it also simply converges.
 
  • #5
sutupidmath said:
Or, i guess we could have also reasoned this way:

SInce |sin(n)|<1 for all n positive integers. then the series

[tex]\sum_{n=1}^{\infty}|sin(n)|^n [/tex] converges, so the original series converges absolutely=> it also simply converges.

I wish we could! :)

think of
[tex]\sum_{n=1}^{\infty} \left( 1 - {1 \over n} \right)^n[/tex]
 
  • #6
nicksauce said:
Based on a maple plot I have made, it appears not to converge.
Plotted is
[tex]f(X) = \sum_{n=1}^{n=X}\sin{n}^n[/tex]
from X=0,100000

The proof I wrote actually shows it DOES NOT converge.

I can't proof it diverges to +infinity yet.

My proof is very tecnical and LONG and actually there are some points a bit difficoult that I have to clear out before to call an EXACT proof.

I need help!
 
  • #7
loop quantum gravity said:
look at exp(in)=cos(n)+isin(n)
(exp(in))^n=exp(in^2)=cos(n^2)+isin(n^2)=(cos(n)+isin(n))^n
Now because exp(in^2) converges also sin(n)^n converges.

I hope I haven't done a mess with the question here.


I can't get why exp(in^2) converges.

Can you explain, please?
 
  • #8
Take_it_Easy said:
I wish we could! :)

think of
[tex]\sum_{n=1}^{\infty} \left( 1 - {1 \over n} \right)^n[/tex]

I wish too!:-p

I wrote before i gave it a thought!:redface:
 
  • #9
Please help me find this simpler limit, no summation. Probably I knew before but not now.

[tex]
\left( 1 + {a \over n} \right)^n
[/tex]
as n goes to infinity, and a is a constant.
 
  • #10
That is a famous limit, you can find it's value by taking a logarithm and using L'Hopital's rule.
 
  • #11
confinement said:
That is a famous limit, you can find it's value by taking a logarithm and using L'Hopital's rule.

:(

(1+1/n)^n:=exp(log(1+a/n)/n)

log(1+a/n)/n=a{log(1+a/n)/(a/n)} -> a{1}=a

since
log(1+x)/x -> log'(1)=1
 
  • #12
[itex]\sum_{n=1}^\infty (sin(n)^n[/itex] diverges.

(please excuse my english, I'm not familiar with math in english, but I hope you'll understand me anyway)

with a sum : [tex]S = \sum^{\infty} u_n[/tex], if S is convergeant, then [tex]\lim_{n \to \infty} u_n = 0[/tex].
by contraposition we can say that if [tex]\lim_{n \to \infty} u_n \ne 0[/tex] then S is not convergeant, and so it is divergeant.

In this case :
[tex]u_n=sin(n)^n[/tex],
and [tex]-1\le sin(n)\le1[/tex], comment : we can't use a strict inegality here
so [tex]-1\le sin(n)^n\le1[/tex],
[tex]\lim_{n \to \infty} u_n \ne 0[/tex],
and so S is divergeant.
 
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  • #13
Actually, you know, I agree with wouzzat. I think he's right... any other thoughts?
 
  • #14
Wouzzat said:
[tex]\[tex]u_n=sin(n)^n[/tex],
and [tex]-1\le sin(n)\le1[/tex], comment : we can't use a strict inegality here
so [tex]-1\le sin(n)^n\le1[/tex],
[tex]\lim_{n \to \infty} u_n \ne 0[/tex],
and so S is divergeant.

Isn't zero (0) also somewhere between -1 and 1? I have forgotten, pardone my bad memory!:-p
 
  • #15
I believe that what he's saying is that no matter how far you go out, it will always keep hitting '1' again every 2PI steps... '-1' every once and a while, it never really approaches zero in any meaningful way.
 
  • #16
csprof2000 said:
I believe that what he's saying is that no matter how far you go out, it will always keep hitting '1' again every 2PI steps... '-1' every once and a while, it never really approaches zero in any meaningful way.

Well, that might be the case, it is just that i don't think we can jumb to that conclusion solely from what is in there, since like i said, 0 is still a possibility to be eventually the limit of that function; weird stuff happens as n-->infty. we need some other way to show that. But you are right, we can see that it does not approach 0 at any finite nr n, but like i said, that is just a sign to prove that it doesn't approach 0, it is not a proof on itself.
 
  • #17
You don't actually go out 2Pi steps in sin(n). In fact, it's fairly easy to see that sin(n) will NEVER be 1 or -1 for n a natural number. The argument doesn't conclude what the limit of sin(n)n as n goes to infinity will be

loop's argument is wrong because ein^2) has absolute value one, so summing over it doesn't converge. In fact, if you can tie in the sin series with the cosine equivalent (cos(n)^n) so that one converges if and only if the other does (not sure if you can though) then this would be a proof that it doesn't converge.

I'd give my solution, but it's super top secret and doesn't exist :p
 
  • #18
Good point, Office Shredder. You're right, of course...
 
  • #19
Probably a dead end but... the question of whether sin(n)n actually goes to 0 can changed into a question of how fast you can approximate pi by rationals (since sin(n) will be close to 1 or -1 when n is close to a multiple of pi, say k*pi, so n/k is close to pi). I distinctly remember something in my algebraic number theory course about algebraic/transcendental numbers being classified based on how quickly you could converge to them based on how large the denominator is, so I'll look that up and see if it helps with anything.

Regardless, I don't think this is the intent of how the problem was supposed to be solved
 
  • #20
sutupidmath said:
Isn't zero (0) also somewhere between -1 and 1? I have forgotten, pardone my bad memory!:-p

sutupidmath said:
we can see that it does not approach 0 at any finite nr n, but like i said, that is just a sign to prove that it doesn't approach 0, it is not a proof on itself.

Hum... yes, actually I didn't proved anything except that "maybe the limit is 0, maybe it isn't".

I'm curious about the final word of that...
 
  • #21
I'm confused. I was taught in my calc class that if the limit of the sequence is not 0 then the series diverges. But someone raised a good point. sin(x) doesn't really have a limit as x[tex]\rightarrow[/tex] infinity. Does that series converge?
 
  • #22
sin(n)^n doesn't converge to 0.

Hint!
 
  • #23
jefswat said:
I'm confused. I was taught in my calc class that if the limit of the sequence is not 0 then the series diverges. But someone raised a good point. sin(x) doesn't really have a limit as x[tex]\rightarrow[/tex] infinity. Does that series converge?

sin(x) doesn't have a limit as x goes to infinity, but sin(n)n might, since raising to the nth power greatly reduces terms less than one. For example, the sequence

1/2,-1/2, 1/2, -1/2,... doesn't converge, but 1/2, (-1/2)2, (1/2)3,... = 1/2,1/4,1/8,1/16,.. does converge
 

Related to Does the Series \(\sum_n (\sin(n))^n\) Converge?

1. Is the series covergent or not?

The answer to this question depends on the specific series and its terms. In order for a series to be convergent, the terms must approach a finite limit as the number of terms increases. If the terms do not approach a finite limit, the series is divergent.

2. How do I determine if a series is convergent or not?

To determine if a series is convergent or not, you can use various tests such as the comparison test, ratio test, or integral test. These tests compare the series to a known convergent or divergent series and can help determine the behavior of the series in question.

3. What is the difference between a convergent and a divergent series?

A convergent series is one in which the sum of the terms approaches a finite limit as the number of terms increases. A divergent series is one in which the terms do not approach a finite limit and the sum of the terms either approaches infinity or oscillates between values.

4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. A series can only exhibit one behavior, either converging to a finite limit or diverging to infinity or oscillating.

5. What are some real-world applications of convergent and divergent series?

Convergent and divergent series have many applications in various fields such as physics, engineering, and economics. For example, in physics, convergent series are used to calculate the value of physical constants and in economics, divergent series are used to model economic growth and inflation.

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