How to determine convergence and divergence

[MHrtz]

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

a_n = 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.

 

[micromass]

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.

 

[MHrtz]

What about other sequences like

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

It's divergent only sometimes right?

 

[micromass]

What about other sequences like

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

It's divergent only sometimes right?

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.

 

[MHrtz]

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

a_n = (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?

 

[micromass]

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 separate 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 separate 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