1. ## convergents of sequence

question: what number does $\displaystyle \frac{2^{n}}{3^{n+1}}$ converge to?
what I've done: sort of use L'Hopital's rule by taking derivative of numerator and denominator => $\displaystyle \frac{n2^{n-1}}{(n+1)3^n}$ so wouldn't this diverge?

2. I think it's easier to just write it as $\displaystyle ~ \frac{1}{3} \cdot \left( \frac{2}{3} \right)^n$

Now, $\displaystyle 0< \frac{2}{3} < 1$, can you see where to go from here?

3. Originally Posted by superdude
question: what number does $\displaystyle \frac{2^{n}}{3^{n+1}}$ converge to?
what I've done: sort of use L'Hopital's rule by taking derivative of numerator and denominator => $\displaystyle \frac{n2^{n-1}}{(n+1)3^n}$ so wouldn't this diverge?
If you wanted to use L'Hospital's rule, you can, because the numerator and denominator both tend to $\displaystyle \infty$.

Your derivatives are also correct, BUT, the numerator and denominator still both tend to $\displaystyle \infty$.

In fact, you would need to work out the numerator's and denominator's derivatives exactly $\displaystyle n$ times to make something not tend to $\displaystyle \infty$.

After $\displaystyle n$ derivatives of the numerator and denominator have been taken, you should have

$\displaystyle \lim_{n \to \infty}{\frac{n\cdot (n - 1)\cdot (n - 2)\cdot\dots \cdot 3\cdot 2\cdot 1 \cdot 2^{0}}{(n + 1)\cdot n \cdot (n - 1) \cdot (n - 2)\cdot \dots \cdot 3 \cdot 2 \cdot 3^1}}$

$\displaystyle = \lim_{n \to \infty}\frac{1}{3(n + 1)}$

$\displaystyle = 0$ since the denominator tends to $\displaystyle \infty$.

Of course, like pomp said, it's easier to just write the fraction as $\displaystyle \frac{1}{3}\cdot \left(\frac{2}{3}\right)^n$.

4. Originally Posted by Prove It
If you wanted to use L'Hospital's rule, you can, because the numerator and denominator both tend to $\displaystyle \infty$.

Your derivatives are also correct, BUT, the numerator and denominator still both tend to $\displaystyle \infty$.
Sorry but when you're taking these derivatives, what are you taking them with respect to?

Surely, $\displaystyle \frac{d}{dn} \left( 2^n \right) = 2^n \log{2}$

5. you're right. thanks
I guess I got Prove It on the wrong track

6. the denominator is tending to infinity faster than the numerator since 3>2, so this sequence converges to zero.

7. Originally Posted by dr.tea
the denominator is tending to infinity faster than the numerator since 3>2, so this sequence converges to zero.
Simpler is the fact that 2/3< 1, as pomp pointed out initially, so the sequence converges to 0.