Sequences and Convergence

  • #1

Homework Statement


Hi everyone. First time trying a forum let alone PhysicsForums.com, everyone seems very nice here.
I am trying to figure out whether a sequence is convergent or not by writing out the first 5 terms. The sequence is: sin[1+(pi/n)]+nsin(pi/n).


Homework Equations


I can't seem to show the limit of the above determined sequence, after finding it is in fact convergent.


The Attempt at a Solution


As I let n=1,2,3,4,5 I got 0.127, 0.0997, 0.0905, 0.0860, 0.0832 respectively.
Now this seems to be heading toward a limit but determining the limit is where I am going wrong, I believe. I sort of cheated and substituted n=1000, 10000 and 1000000 and got 0.07233, 0.07228 and 0.07228 again from the graphics calculator.
Now I believe the limit is approx. 0.0722... but how could I possibly go about finding this from the original sequence?
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,783
18
Is there some restriction on the tools you can use?

Cuz it's easy with l'Hospital's rule.
 
  • #3
There is no restriction in how I could show the limit. But doesn't L'Hopital's rule only apply to fractions e.g. f(x)/g(x)=f'(x)/g'(x). How could it be applied for this example?
 
Last edited:
  • #4
quasar987
Science Advisor
Homework Helper
Gold Member
4,783
18
cuz n=1/(1/n)
 
  • #5
n=1/(1/n)? I'm not sure where this came from. How did a value for n cone about and do I substitute it back into sin[1+(pi/n)]+nsin(pi/n)?
Sorry quasar987 I just don't understand how that works.
 
  • #6
1,631
4
As long as i can see, using quasar's approach , and after that the l'hopital rule, the liimit is going to be 0.
 
  • #7
Ok but I just can't seem to understand quasar's approach. How did he get the value of n to be 1/(1/n)? And do I apply L'Hopitals rule to n and then substitute back?
 
  • #8
1,631
4
Ok but I just can't seem to understand quasar's approach. How did he get the value of n to be 1/(1/n)? And do I apply L'Hopitals rule to n and then substitute back?


Notice that n= 1/(1/n), he did nothing here, it is just a double fraction

like doing 2= 1/(1/2), you see it is the same.
 
  • #9
VietDao29
Homework Helper
1,423
3
...
I am trying to figure out whether a sequence is convergent or not by writing out the first 5 terms. The sequence is: sin[1+(pi/n)]+nsin(pi/n).

What quasar987 meant is to change n to 1 / (1 / n). Since you need a fraction, and you only have a multiplication, so what you should do is to change multiplication to division.

There are several ways to go about solving this problem. L'Hopital rule is one way to go, however, it a little bit overkill. It goes like this:
[tex]\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}[/tex]

Now, can you apply L'Hopital to this problem?

--------------------------

Another nicer way is to use the well-know limit:
[tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]
The first few steps are just the same:

[tex]\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}[/tex]

As [tex]n \rightarrow \infty[/tex], we have [tex]\frac{1}{n} \rightarrow 0[/tex], right? Let t = 1 / n, so your limit becomes:

[tex]... = \sin 1 + \lim_{t \rightarrow 0} \frac{\sin (\pi t)}{t}[/tex], now, you'll try to rearrange your limit so that it would look the same as:
[tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]

Multiply both numerator, and denominator by pi, yields:
[tex]... = \sin 1 + \pi \lim_{t \rightarrow 0} \frac{\sin (\pi t)}{\pi t}[/tex]

You should be able to go from here, right? :)
 
  • #10
Ah I see now, cheers guys, especially VietDao29 for that thorough and comprehensive explanation.
 

Related Threads on Sequences and Convergence

Replies
7
Views
7K
  • Last Post
Replies
5
Views
2K
  • Last Post
2
Replies
31
Views
3K
  • Last Post
Replies
2
Views
978
Replies
4
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
5
Views
10K
Replies
4
Views
6K
  • Last Post
Replies
2
Views
613
  • Last Post
Replies
14
Views
1K
Top