Limit of $\frac{n^{3}}{2^{n}}$

Question

Compute the Limes for $n\rightarrow\infty$ for the sequence:

$c_{n} :=\frac{n^{3}}{2^{n}}, for\ n\in N_{0}$

I know the limit of this sequence is 0. Just because the limit of any polynomial divided by an exponential function will always diverge to 0.

But since we have to actually compute it, I would like to ask for some help.

Since we haven't had derivatives in class I cannot simply use L'Hopital( neither can I use an e-function or the natural log).

Answer 1

Proof from these four facts:
(A) Bernoulli's inequality
$(1+a)^n \ge 1+an,$ for $a>-1$, $n=1,2,3\dots$. Proved by induction on $n$.

(B) If $a_n$ is positive and eventually decreasing, and some subsequnce converges to $0$, then $a_n$ also conveges to $0$.

(C) $\frac{1}{n} \to 0$ as $n \to \infty$. Proved from the archimedean property for the reals: i.e., the set of positive integers $\mathbb N$ is not bounded above.

(D) If $a_n \to 0$ and $b_n \to 0$, then $a_n b_n \to 0$.

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Claim 1. $\frac{1}{2^n} \le \frac{1}{n}$.
Proof. From Bernoulli, $2^n = (1+1)^n \ge 1+n$, so $2^n \ge n$, and finally $\frac{1}{2^n} \le \frac{1}{n}$.

Claim 2. $\frac{1}{2^n} \to 0$.
Proved from Claim 1 and (C).

Claim 3. $\frac{n^3}{2^n} \to 0$.
Proof. The seqeuce is decreasing for $n \ge 4$. By (B), it is enough to prove that the subsequence $\frac{(4n)^3}{2^{4n}}$ converges to $0$. Compute:

\begin{align} \frac{(4n)^3}{2^{4n}} &= 4^3 \cdot \frac{n}{2^n}\cdot\frac{n}{2^n}\cdot \frac{n}{2^n}\cdot\frac{1}{2^n} \\ &\le 4^3 \cdot 1\cdot 1\cdot 1\cdot \frac{1}{2^n} \qquad\text{by Claim 1} \\ &\to 4^3 \cdot 1\cdot 1\cdot 1\cdot 0 \qquad\text{by Claim 2 and (D)} \\ &= 0 . \end{align} Therefore, $\frac{(4n)^3}{2^{4n}} \to 0$, and thus $\frac{n^3}{2^n} \to 0$ by (B).

Answer 2

To prove that the limit is $0$, you need to choose a series of values $\varepsilon$ converging to $0$ and some integers $N(\varepsilon)$ depending on $\varepsilon$ for which you hope to show that $|c_n| < \varepsilon$ for any $n > N(\varepsilon)$, then check that this is indeed the case.

I recommend the choice of $\varepsilon = 2^{-k}$ and $N = 2^k$ for all integers $k > 4$.

Answer 3

$$\frac{c_{n+1}}{c_n}=\frac12\left(1+\frac1{n}\right)^3.$$

So for all

$$n\ge4>\frac1{\sqrt[3]2-1}$$ this ratio is no greater than $r:=\dfrac{125}{128}<1$. So

$$c_n\le c_4r^{n-4}.$$

Answer 4

Using the binomial theorem, we have $$2^n = \sum_{k=0}^n {n \choose k} \geq \frac{n(n-1)(n-2)(n-3)}{24},$$ for $n$ sufficiently large. (For $n \geq 4$ to be precise, the term has to show up in the binomial expansion, of course.) So \begin{equation} 0 \leq \frac{n^3}{2^n} \leq \frac{24n^3}{n(n-1)(n-2)(n-3)} = \frac{24}{n -6 + 11n^{-1} - 6n^{-2}}, \label{x} \end{equation} for $n$ sufficiently large. The denominator of this fraction goes to $+\infty$ as $n \to \infty.$ The numerator is bounded. Therefore the expression on the right hand side of the inequality above goes to $0$ as $n \to \infty.$ Using the squeeze theorem, we obtain $\frac{n^3}{2^n} \to 0$ as $n \to \infty.$

(It's a very good exercise for students new to limits of sequences to show in general that if $(b_n)_n$ goes to $\pm \infty$ and $(a_n)_n$ is bounded that $(a_n/b_n)_n$ goes to zero.)

Answer 5

Root test:

$a_n:=(c_n)^{1/n}=(n^{1/n})^3/2$;

$\lim_n a_n=1/2<1,$

the series $\Sigma c_n$ converges, hence

$\lim_n c_n=0.$

Answer 6

Use a "telescoping product trick":

For each integer $n,\ \frac{(n+1)^3}{n^3}=1+\frac{3}{n^2}+\frac{3}{n}$ so $d_n:=\frac{c_{n+1}}{c_n}=\left(1+\frac{3}{n^2}+\frac{3}{n}\right )\cdot \frac{1}{2}.$

Now compute that

$c_{n+1}=2\prod_{k\:=\:1}^n\ \frac{d_{k+1}}{d_k}\le (2)(7)\frac{1}{2^n}=\frac{14}{2^n}\to 0$ as $n\to \infty.$