$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$

Question

How do I solve this example? I tried to point out the fastest growing term $2 ^ n$ and $3 ^ n$, but that doesn't seem to lead to the result. I know the limit is $0$ that's obvious, but I don't know how to work on it.

$$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$$

Or can I make a power estimate of a theorem on a tightened sequence, for example, that $2 ^ n$ / $13 ^ n$ (smaller) goes to zero and the other side $2 ^ n$ / $3 ^ n$ (bigger) also goes to zero?

$\lim_{n\to\infty} 2 ^ n$ / $13 ^ n$< $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ < $\lim_{n\to\infty} 2 ^ n$ / $3 ^ n$

Answer 1

It follows from the ratio test.

If $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = a < 1 \ $, then $\lim a_n = 0$.

$$ \dfrac{\frac{2^{n+1} \left((n+1)^4-1\right)}{4\cdot3^{n+1} + (n+1)^7}}{\frac{2^n(n^4-1)}{4\cdot3^n + n^7}} = \dfrac{2^{n+1} \left((n+1)^4-1\right)(4\cdot3^n + n^7)}{2^n(n^4-1)\left(4\cdot3^{n+1} + (n+1)^7\right)} \\ =\frac{2}{3} \left(\frac{n+1}{n}\right)^4 \left(\dfrac{1-\frac{1}{(n+1)^4}}{1-\frac{1}{n^4}}\right) \left(\dfrac{1+\frac{n^7}{3^n}}{1+\frac{(n+1)^7}{3^{n+1}}} \right) \longrightarrow \frac{2}{3} < 1 $$

Answer 2

Observe that $$ f(n)=\frac{(\frac23)^n\cdot (n^4-1)}{4+(\frac13)^n\cdot n^7} $$

If you know the following limits, you are done: $$ \lim_{n\to\infty} (\frac23)^nn^4=0,\quad \lim_{n\to\infty} (\frac23)^n=0, \quad \lim_{n\to\infty}(\frac13)^n\cdot n^7=0 $$

All you need is the following general statement:

If $0<r<1$ and $\alpha$ is a real number, then $$ \lim_{n\to\infty}{n^\alpha}r^n $$

To see why, let $k>0$ be an integer such that $k>\alpha$. Then for $n>2k$, $$ (1+p)^n>\binom{n}{k}p^k=\frac{n(n-1)\cdots(n-k+1)}{k!}p^k>\frac{n^kp^k}{2^kk!},\qquad r=:\frac{1}{1+p} $$ So $$ 0<n^\alpha r^n<\frac{2^kk!}{p^k}n^{\alpha-k},\quad (n>2k) $$ Since $\alpha-k>0$, $n^{\alpha-k}\to 0$.

Answer 3

This is only a matter of considering the infinite hierachy: at the numerator the term $2^n$ is the fastest to go to $+\infty$ and at the denominator is $3^n$. So the resulting limit to determine is: $\lim_{n\to\infty}\frac{2^n}{4\cdot 3^n}=\lim_{n\to\infty}\frac{1}{4}(\frac{2}{3})^n=0$, (this is $0$ since $\lim_{n\to\infty}a^n=0$ if $|a|<1$).