Determine if sequence converges or diverges

[MarkSA]

Hello,

I had a problem like such:

Determine if the sequence is convergent or divergent. If it's convergent find the limit.

a(sub n) = 3^(n)/[2^(2n + 1)]

Is this a problem you would have to solve by plugging in values for a(sub 1), a(sub 2), etc and observing what occurs as n->infinity? Or is there a more 'correct' (but fairly easy) way of doing it, like with limits? 'Guessing' this way I came up with the limit being = 0 convergent.

I tried taking the limit of it as n->infinity since that would tell me about convergence/divergence.
lim as n-> infinity of: 3^(n)/[2^(2n + 1)]
But it appears to be indeterminate in that form, infinity/infinity. So I thought about using the Hospital rule. However, the derivative of a^(x) is a^(x) * ln(a) I believe. I can't see that getting me anywhere since i'll still end up with an infinity in numerator and denominator thanks to an a^(x).

How else could this problem be solved?

 

[pka]

\(\displaystyle \frac{{3^n }}{{2^{2n + 1} }} = \frac{1}{2}\left( {\frac{3}{4}} \right)^n\)

 

[MarkSA]

Ok thanks, I see it now.

I keep getting taken for a ride on these problems that involve manipulating the exponents. Hopefully once I see enough examples though I will be able to recognize them...

Part of it was my calc TA had mentioned that if it was n +/- a number, it was doable (I have those down now hopefully).. but he said if it's n multiplied by something else there's nothing you can do. That kind of misled me. I wonder if he was referring to something else about these problems where 2n or 3n etc makes it impossible.