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?
Thanks!
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?
Thanks!
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.
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 $n\to\infty$.
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.