# Sequences and Convergence

## 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?

quasar987
Homework Helper
Gold Member
Is there some restriction on the tools you can use?

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

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:
quasar987
Homework Helper
Gold Member
cuz n=1/(1/n)

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.

As long as i can see, using quasar's approach , and after that the l'hopital rule, the liimit is going to be 0.

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?

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.

VietDao29
Homework Helper
...
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:
$$\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}$$

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? :)

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