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How to prove that exponential grows faster than polynomial?

I have this sequence with $b>1$ and $k$ a natural, which diverges: $\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty$ I need to prove this, with what i have learnt till now from my textbook, my simple step is this:

Since $n^2\leq2^n$ for $n>3$, i said $b^n\geq n^k$, so it diverges. Is it right?

I am asking here not just to get the right answer, but to learn more wonderful steps and properties.

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    Only your instructor knows what you have done in your book, so why not ask him/her?2012-12-28

5 Answers 5

0

Apply L'Hospital's Rule $k$ times to get $\frac{(\log b)^k b^n}{k!}$ if $k>0$

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    May I expect a comment on fault by the down-voter?2012-12-28
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Note that $b^n = \exp(n \log b) = \displaystyle \sum_{j=0}^{\infty} \dfrac{(n \log b)^j}{j!} \geq \dfrac{(n \log b)^{k+1}}{(k+1)!}$ Hence, we have $\dfrac{b^n}{n^k} \geq \dfrac{(n \log b)^{k+1}}{(k+1)! n^k} = \dfrac{(\log b)^{k+1}}{(k+1)!} n$ Hence, $\lim_{n \to \infty} \dfrac{b^n}{n^k} \geq \lim_{n \to \infty} \dfrac{(\log b)^{k+1}}{(k+1)!} n = \infty$

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    @johnny Thanks. I have changed the exponent from $k$ to $k+1$.2012-12-28
4

$\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty$

You can use the root test, too: $\lim_{ n\to \infty}\sqrt[\large n]{\frac{b^n}{n^k}} = b>1$

Therefore, the limit diverges.


The root test takes the $\lim$ of the $n$-th root of the term: $\lim_{n \to \infty} \sqrt[\large n]{|a_n|} = \alpha.$

If $\alpha < 1$ the sum/limit converges.

If $\alpha > 1$ the sum/limit diverges.

If $\alpha = 1$, the root test is inconclusive.

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    great, thanks dude2012-12-28
2

We have $b^n/n^k=\exp(n\ln b-k\ln n)=\exp(n\ln b(1-\frac{k}{\ln b}\frac{\ln n}{n}))$.

Now $\lim_{n\rightarrow +\infty} n\ln b=+\infty$ and $\lim_{n\rightarrow +\infty} 1-\frac{k}{\ln b}\frac{\ln n}{n}=1$.

So $\lim_{n\rightarrow +\infty}n\ln b(1-\frac{k}{\ln b}\frac{\ln n}{n})=+\infty$.

It follows that $\lim_{n\rightarrow +\infty} b^n/n^k=+\infty$.

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    nice, thanks Julien2012-12-28
2

This follows immediately from Bernoulli:

$\frac{b^n}{n^k}= (\frac{b^{\frac{n}{2k}}}{\sqrt{n}})^{2k}$

Now, by Bernoully

$b^{\frac{n}{2k}} \geq 1+\frac{n}{2k}(b-1)$

Thus $\frac{b^n}{n^k} \geq (\frac{1}{\sqrt{n}}+\frac{\sqrt{n}}{2k}(b-1))^{2k} \geq (\frac{\sqrt{n}}{2k}(b-1))^{2k}$

Since th RHS goes to infinity, you are done.