# How to determine convergence and divergence

1. Jun 30, 2011

### MHrtz

I've been having some trouble understanding how to determine if a sequence is divergent or convergent. For example

an = cos(2/n)

I know if I take the limit as n ->$\infty$ then I will get 1. So the sequence has a limit but does having a limit mean that the sequence is convergent.

2. Jun 30, 2011

### micromass

Staff Emeritus
Hi MHrtz!

Yes, saying that a sequence has a limit (which should not be infinity) is equivalent to saying that the sequence converges.
If the limit doesn't exist (or is infinite), then the sequence diverges.

3. Jun 30, 2011

### MHrtz

an {1 n = 2k k is an integer
{0 otherwise

It's divergent only sometimes right?

4. Jun 30, 2011

### micromass

Staff Emeritus
A sequence is either convergent or divergent. There's no such thing as divergent "only sometimes". This particular sequence is divergent since it doesn't come close to any number.

5. Jun 30, 2011

### MHrtz

Ok, so i did some more problems and came across this one:

an = (1 + 2/n)n

When I took the limit I though it was 1 but the book said that the limit was e2. How is this possible?

6. Jun 30, 2011

### micromass

Staff Emeritus
The problem is that you probably reason as follows:

(1+2/n) goes to 1, n goes to infinity, so (1+2/n)n goes to $1^{+\infty}=1$. The problem is however that $1^{+\infty}$ is an undetermined form and does not equal 1!! Check your calculator, if you type in large values of n, then you'll see that (1+2/n)n does not go to 1!!

How to solve this problem then. Well, it depends on what you seen.
Some students have $(1+1/n)^n\rightarrow e$ as a seperate formula. Then you just need to transform (1+2/n)n into something of the form (1+1/m)m (hint: m=n/2)

If you haven't seen that seperate formula, then you can always do

$$(1+2/n)^n=e^{n\log(1+2/n)}$$

so you just need to show that

$$n\log(1+2/n)\rightarrow 2$$