Limit of Sequence a_n: Find Limit as n->∞

[merced]

a_n = 2^n/(3^n + 1)

Find the limit as n -> infinity

So I separated the fraction into (1/3)(2/3)^n

Then limit = 1/3 lim_(n->infinity) (2/3)^n.

So, my question is, how do you find the limit (2/3)^n. Is analysis of functions simply enough, i.e. looking at how fast they increase?

 

[quasar987]

How did you transform a_n into (1/3)(2/3)^n?

Btw - is this for a real analysis class or some intro calculus class?

 

[merced]

Oops, sorry, I guess I didn't look hard enough at my post.

It's really a_n = \frac{2^n}{3^{n+1}}

And it's Calculus II

 

[quasar987]

Oh ok.

Didn't you see the general result that if -1<a<1, then a^n\rightarrow 0?

If not, and if you've seen a little bit of series theory, you can prove it this way: You know that the geometric series \sum a^n converges for -1<a<1 and diverges otherwise. And you've also seen the basic necessary criterion for a series to converge, namely that the general term a_n of any converging series goes to zero in the limit n-->infty.

So you can make use of these two fact by saying "I know that the geometric series of general term a_n=a^n (where -1<a<1) converges, hence it must be that a_n=a^n\rightarrow 0 for -1<a<1."

 

[merced]

Ok, I get what you're saying...but why is -1<a<1 here?

 

[quasar987]

I proved that for any real number a btw -1 and 1, the limit of a^n is zero.

In particular, for a=2/3, we have that the limit of (2/3)^n is zero. Which is what you were wondering about.

 

[merced]

Ooh, ok, I understand now.

Thanks!