# Homework Help: Sequences and Convergence or Divergence?

1. Oct 2, 2007

### Illusionist

1. The problem statement, all variables and given/known data
Determine whether the following sequence, whose nth term is given, converges or diverges. Find the limit of each convergent one.

n[1 - cos(2/n)]

2. Relevant equations
I have made a solid attempt and obtained an answer but I am convinced I made a mistake and have missed something.

3. The attempt at a solution
lim n[1 - cos(2/n)]
lim [ n - n[cos(2/n)] ]
lim(n) - lim (n[cos(2/n)])
lim(n) - lim ( [cos(2/n)] / (1/n) ) <--- Now using L'Hopital's Theorem:
lim(n) - lim ( [(-2/n^2)sin(2/n)] / (-1/n^2) )
lim(n) - lim 2sin(2/n)
lim(n) - 2sin(lim [2/n])
Infinity - 0 = Infinity, hence divergent!

Note: Each limit (lim) shown about is the limit as n approaches infinity, just didn't write it every time to make things a bit neater.

Now for some reason that seems to simple and incorrect to me. Confirmation or any sort of help would be much appreciated.
Sorry about the confusing working above, thanks guys.

2. Oct 2, 2007

### NateTG

You can't apply l'Hospital's theorem to:
$$\frac{\cos (\frac{2}{n})}{\frac{1}{n}}$$
since it's not of a suitable form. (The numerator goes to 1.)

3. Oct 2, 2007

### Illusionist

Of course but I'm a bit stumped as to what to do with this one. Any suggestions? I presume that means what I did before applying L'Hopital's was right. Thanks for the reply.

4. Oct 2, 2007

### Dick

$$\frac{1-\cos (\frac{2}{n})}{\frac{1}{n}}$$ is in a good form to apply L'Hopital to, isn't it?

5. Oct 2, 2007

### rock.freak667

if $cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$ replace all the x's with 2/n and you'll get the series expansion for cos(2/n)

then simplify $n(1-cos(\frac{2}{n}))$ and then check what happens as n tends to infinity....

I did this just to make the 1 in the problem disappear

6. Oct 3, 2007

### Illusionist

Very true. Well I gave that a try and this is how I went:

lim [(-2/n^2)sin(2/n)] / (-1/n^2)

lim 2sin(2/n)

2sin (lim [2/n])

2sin(0)

0, hence convergent.

Please tell me I'm right! Thank you so much for the replies guys.

7. Oct 3, 2007

### Dick

That's what I would do. And that's the answer I would get.

8. Oct 3, 2007

### dynamicsolo

There's nothing wrong with your method, but if Illusionist is in a typical calculus course introducing infinite series, you are anticipating things a little. Generally, convergence of infinite series and problems of the sort posted appear a while before Taylor series are discussed. (Otherwise, yeah, Taylor series certainly are a "nice big hammer" for dealing with problems of this sort...)