Exercise II.2.2: Prove that $\lim \dfrac{n^3}{2^n} = 0$ I'm trying to solve Problem II.2.2 in textbook Analysis I by Amann/Escher.



Could you please verify whether my attempt on (b) and (c) is fine or it contains logical gaps/errors?

My attempt:
From binomial theorem, we have $$2^n = (1+1)^n = \sum_{k=0}^n {n \choose k} = {n \choose 4} + \sum_{k=0 \atop k\neq 4}^n {n \choose k} = n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}$$
As a result, $$\begin{aligned} \lim \dfrac{n^3}{2^n} &= \lim \dfrac{n^3}{n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}} \\ &= \lim \dfrac{1}{n \left( 1-\dfrac{1}{n} \right) \left(1-\dfrac{2}{n} \right) \left(1-\dfrac{3}{n} \right) + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} \\ &= \lim \dfrac{1}{n + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} = 0\end{aligned}$$

Your idea is correct, but you can be more concrete and more simple:
It is $2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}>\binom{n}{4}$ for $n$ big enough.
$\binom{n}{4}=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{24}<\frac{n^4}{24}$
We can stipulate like this:
$\lim_{n\to\infty} \frac{n^3}{2^n}\leq\lim_{n\to\infty} \frac{n^3}{n^4/24}=\lim_{n\to\infty} \frac{24}{n}=0$
Because you have certainly already proven, that $\lim_{n\to\infty} \frac1n=0$
