Help with a limit

**Cakecake** · January 6th 2009, 05:58 PM

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

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

**Mathstud28** · January 6th 2009, 06:09 PM

Originally Posted by Cakecake

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

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

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

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

This implies that

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

Now you should know that if $\beta>0$ that $\lim_{n\to\infty}\frac{1}{n^{\beta}}=0$ ...so since $\alpha-K<0$ this proves that $\lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$ ...take $\alpha=0,p=\frac{1}{3}$ 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 $\left|\frac{3}{4}\right|<1$ so $\lim_{n\to\infty}\left(\frac{3}{4}\right)=0$

**Cakecake** · January 6th 2009, 06:20 PM

Woah I was looking for a much simpler answer.

Thanks anyway for typing all that up.

**Mathstud28** · January 6th 2009, 06:24 PM

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 $\left|\frac{3}{4}\right|<1$ that $\lim_{n\to\infty}\left(\frac{3}{4}\right)^n=0$ . What level are you currently at?

**Cakecake** · January 6th 2009, 06:31 PM

I guess that could work.

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

**Mathstud28** · January 6th 2009, 06:35 PM

Originally Posted by Cakecake

I guess that could work.

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

Perfect

! Consider that $\lim_{n\to\infty}\sqrt[n]{\left(\frac{3}{4}\right)^n}=\frac{3}{4}<1\implies \sum_n \left(\frac{3}{4}\right)^n\text{ converges}$ . Now what is one of the neccessary conditions for series convergence?

**Cakecake** · January 6th 2009, 06:45 PM

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

**Mathstud28** · January 6th 2009, 06:47 PM

Originally Posted by Cakecake

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

...well there you go

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