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I can think of a number of ways to prove that $\displaystyle\lim_{n\rightarrow \infty} (4^{n}-n^{4})= \infty$ but none of them brief. Does anyone have a short proof of this?


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let's see one or two. – jspecter Oct 15 '11 at 19:49
One: I won't write the entire thing, but for example you can convert both parts of the difference to $e^{x}-e^{y}$ form and use the fact e is monotone increasing. Then you need to prove the difference of $x-y$ approaches infinity, which is also simple but pretty long (I took derivatives to show that), and finally use the fact that the limit of $e^{n}$ is infinity. – rowel Oct 15 '11 at 19:58
Two: first prove the expression $ln(4^{n}-n^{4})$ is defined beyond n>100, then use $ln$ on the expression and do mostly the same thing as in the first proof. For such a simple limit, both of these proofs look too complex to me! – rowel Oct 15 '11 at 20:05
up vote 3 down vote accepted

The second term is very small compared to the first as $n$ gets large. Specifically $$\lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n} = \lim_{n \rightarrow \infty} (1 - {n^4 \over 4^n})$$ $$= 1 - \lim_{n \rightarrow \infty} {n^4 \over 4^n}$$ The limit on the right can be shown to be zero in several ways, such as applying L'hopital's rule several times. So the ratio goes to $1$ as $n$ goes to infinity. As a result, $$\lim_{n \rightarrow \infty} 4^n - n^4 = \lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n} 4^n$$ $$= \lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n}\lim_{n \rightarrow \infty} 4^n$$ $$= \infty$$ (The product rule for limits works for such limits).

Alternatively, the above limit shows that for $n$ large enough you have $${4^n - n^4 \over 4^n} > {1 \over 2}$$ This means $$4^n - n^4 > {1 \over 2}4^n$$ So since $\lim _{n \rightarrow \infty} 4^n = \infty$, $\lim_{n \rightarrow \infty} 4^n - n^4 = \infty$ as well.

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the problem is to find the limit for (4^n - n^4). There is no 4^n in the denominator. – NoChance Oct 15 '11 at 20:37
This is so simple, why did it slip my mind? I think this is the answer I'll accept - but I'll wait for an hour or so since there seem to be different approaches to this. Thank you. – rowel Oct 15 '11 at 20:37
@EmmadKareem I know, I am just using the limit of the ratios to find the limit he seeks: If $f/g \rightarrow 1$ and $g \rightarrow \infty$, then $f \rightarrow \infty$. (You can use product rules for limits if you like: $f = (f/g)*g$. – Zarrax Oct 15 '11 at 20:42
@rowel: I guess you should be more explicit which things are you allowed to use to find simple limit as $4^n-n^4$. I agree with Emmad that working with these limits is quite equivalent – Ilya Oct 15 '11 at 20:44
@Gortaur: as this isn't a homework problem I'm allowed to use 'all I know', which is almost nil about limits of series and high school-level about limits of functions in $R$, but you are right I should've said so. I will be sure to mention this next time, thank you. – rowel Oct 15 '11 at 20:47

Denote $a_n = 4^n$ and $b_n = n^4$. Clearly, $a_4 = b_4$ and $a_6>2b_6$. Suppose that for some $n\geq 6$ it holds $a_n\geq 2b_n$ then $$ a_{n+1}=4a_n\geq 8b_n=8\left(\frac{n}{n+1}\right)^4b_{n+1}\geq 2b_{n+1} $$ and by induction you have $a_n\geq 2b_n$ so $a_n-b_n\geq b_n\to\infty$ with $n\to\infty$.

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Consider the following form of your limit:

$$\lim_{n\rightarrow \infty} (4^{n}-n^{4})=\lim_{n\rightarrow \infty} (2^{n}-n^{2})(2^{n}+n^{2})\geq \lim_{n\rightarrow \infty}\left(2\dbinom{n}{1}+2\dbinom{n}{2} -n^{2}\right)(2^{n}+n^{2})=\lim_{n\rightarrow \infty}n(2^{n}+n^{2}) \longrightarrow \infty$$

Above i used the fact that: $$\sum_{k=0}^{n} \dbinom{n}{k}=2^n$$ Hence, the limit is $\infty$.

The proof is complete.

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