# prove $\lim_ {n \rightarrow \infty} \frac{b^n}{n^k}=\infty$ [duplicate]

Possible Duplicate:
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.

-

## marked as duplicate by Jonas Meyer, TMM, Davide Giraudo, rschwieb, Ittay WeissDec 28 '12 at 20:37

(1) It is not clear how “$n^2 \leq 2^n$” implies “$b^n \geq n^k$”. Note that the latter is true only for sufficiently large $n$ (for every $b>1$ there exists $N$ s.t. for $n\geq N$ we have $b^n \geq n^k$). (2) The fact that $b^n \geq n^k$ doesn't imply that $b^n/n^k \to \infty$ as $n\to\infty$. – Yury Dec 28 '12 at 19:24
@Jonas, it is not duplicate, the other question is just the contrary – doniyor Dec 28 '12 at 19:30
@doniyor: I do not know what you mean. It is essentially a duplicate, in my opinion, but it has already been answered here, too, and will only be closed as a duplicate if 4 others with enough points agree. For a positive sequence $(a_n)$, note that $a_n\to+\infty$ if and only if $\dfrac{1}{a_n}\to 0$. – Jonas Meyer Dec 28 '12 at 19:34
@Jonas, Okay, thanks buddy – doniyor Dec 28 '12 at 19:35
Only your instructor knows what you have done in your book, so why not ask him/her? – GEdgar Dec 28 '12 at 19:54

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

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.

-
great, thanks dude – doniyor Dec 28 '12 at 19:32

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, $$\lim_{n \to \infty} \dfrac{b^n}{n^k} \geq \lim_{n \to \infty} \dfrac{(\log b)^{k+1}}{(k+1)!} n = \infty$$

-
There's a $+1$ missing in the first line inequality. Nice idea though. – johnny Dec 28 '12 at 19:46
@johnny Thanks. I have changed the exponent from $k$ to $k+1$. – user17762 Dec 28 '12 at 19:47

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

-
nice, thanks Julien – doniyor Dec 28 '12 at 19:35

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.

-

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

-
i have not come to L'Hospitals rule in my book – doniyor Dec 28 '12 at 19:24
Are you going to the hospital? – TMM Dec 28 '12 at 19:25
Maybe he knows the Marquis lived before French spelling reform. – GEdgar Dec 28 '12 at 19:52
May I expect a comment on fault by the down-voter? – lab bhattacharjee Dec 28 '12 at 20:22