# Sequences and Convergence

1. May 12, 2007

### SoonerTheBetter

1. The problem statement, all variables and given/known data
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).

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

3. 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?

2. May 12, 2007

### quasar987

Is there some restriction on the tools you can use?

Cuz it's easy with l'Hospital's rule.

3. May 12, 2007

### SoonerTheBetter

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: May 12, 2007
4. May 13, 2007

### quasar987

cuz n=1/(1/n)

5. May 13, 2007

### SoonerTheBetter

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. May 13, 2007

### sutupidmath

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. May 13, 2007

### SoonerTheBetter

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. May 13, 2007

### sutupidmath

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. May 13, 2007

### VietDao29

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:
$$\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}$$

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

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

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

$$\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}$$

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

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

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

You should be able to go from here, right? :)

10. May 14, 2007

### SoonerTheBetter

Ah I see now, cheers guys, especially VietDao29 for that thorough and comprehensive explanation.