Thread: Help with a limit

1. Help with a limit

$\displaystyle \lim_{n \to \infty} (\frac{3}{4})^n$

I know the answer is 0. But exactly how can I show that?

2. Originally Posted by Cakecake
$\displaystyle \lim_{n \to \infty} (\frac{3}{4})^n$

I know the answer is 0. But exactly how can I show that?
By prove it what tools do you mean? By delta-epsilon defintion?

We can prove that if $\displaystyle p>0,\alpha\in\mathbb{R}$ then $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$

Proof: Consider an integerl $\displaystyle K>\alpha,K>0$. For $\displaystyle n>2K$

$\displaystyle (1+p)^n> {n\choose {K}} p^K>\frac{n^Kp^K}{2^KK!}$

This implies that

$\displaystyle 0\leqslant\frac{n^{\alpha}}{(1+p)^n}\leqslant\frac {2^KK!}{p^K}n^{\alpha-K}~~~n>2K$

Now you should know that if $\displaystyle \beta>0$ that $\displaystyle \lim_{n\to\infty}\frac{1}{n^{\beta}}=0$...so since $\displaystyle \alpha-K<0$ this proves that $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$...take $\displaystyle \alpha=0,p=\frac{1}{3}$ to acheive your limit.

Is that what you wanted? Or were you looking for something simpler?

Note: I think if you weren't looking for a proof like the one above you could just state that $\displaystyle \left|\frac{3}{4}\right|<1$ so $\displaystyle \lim_{n\to\infty}\left(\frac{3}{4}\right)=0$

3. Woah I was looking for a much simpler answer.

Thanks anyway for typing all that up.

4. Originally Posted by Cakecake
Woah I was looking for a much simpler answer.

Thanks anyway for typing all that up.
Then I think it would suffice to say that since $\displaystyle \left|\frac{3}{4}\right|<1$ that $\displaystyle \lim_{n\to\infty}\left(\frac{3}{4}\right)^n=0$. What level are you currently at?

5. I guess that could work.

I'm taking BC, currently a junior. I'm up to series.

6. Originally Posted by Cakecake
I guess that could work.

I'm taking BC, currently a junior. I'm up to series.
Perfect ! Consider that $\displaystyle \lim_{n\to\infty}\sqrt[n]{\left(\frac{3}{4}\right)^n}=\frac{3}{4}<1\implies \sum_n \left(\frac{3}{4}\right)^n\text{ converges}$. Now what is one of the neccessary conditions for series convergence?

7. The limit of it as n goes to infinity has to be 0.

8. Originally Posted by Cakecake
The limit of it as n goes to infinity has to be 0.
...well there you go