Prove the existence of an $n'$ such as for all $n>n'$ : $\frac{U_{n+1}}{U_n} < 3/4$

Question

Let $U_n$ be a numerical sequence defined by $U_n=n^2/2^n$ We need to deduce from the fact that the limit of $U_{n+1}/U_n$ equals $1/2$ the existence of $n'.$ How am I supposed to prove its existence. It's clear that next to $+\infty$ this condition is valid ; Am I right?

Answer 1

Hint In the definition of limit use $\epsilon =\frac{1}{4}$.

Answer 2

It is $$\frac{U_{n+1}}{U_n}=\frac{(n+1)^2\times 2^n}{2^{n+1}\times n^2}=\frac{(n+1)^2}{2n^2}<\frac{3}{4}$$ and this is equivalent to $$4(n+1)^2<6n^2$$ Can you finish?

Answer 3

By the definition of the limit $\displaystyle \lim_{n\to +\infty}a_n=a$, for every $\varepsilon>0$ there is $n'\in \mathbb{N}$ such that for all $n>n'$ we get $|a_n-a|<\varepsilon.$ Since you may easily deduce that $\frac{U_{n+1}}{U_n}\to \frac{1}{2}$ (for example by the algebra of limits), you know that:

for every $\varepsilon>0$ there is $n'\in \mathbb{N}$ such that for all $n>n'$ you have $\left|\frac{U_{n+1}}{U_n}-\frac{1}{2}\right|<\varepsilon.$

Whenever in Math you know something for every quantity, say $A$, then it is very probable that you need to use a specific (yes, specific!) $A$ in order to obtain a specific result. Like here: by setting $\varepsilon$ to be $\frac{1}{4}$ (you shall see why) you obtain $n'\in \mathbb{N}$ such that for all $n>n'$ you have: $$\left|\frac{U_{n+1}}{U_n}-\frac{1}{2}\right|<\frac{1}{4}\iff -\frac{1}{4}<\frac{U_{n+1}}{U_n}-\frac{1}{2}<\frac{1}{4},$$ that is: $$\frac{1}{4}<\frac{U_{n+1}}{U_n}<\frac{3}{4},$$ your desired result.

Note that $\varepsilon=\frac{1}{4}$ does the job, but it is not the only one: anything less that you choose, would do the job also!

Answer 4

$$\left\{ \begin{array}{l} \frac{U_{n+1}}{U_n}<\frac34 \\[2ex] U_n=\frac{n^2}{2^n} \end{array} \right) \Rightarrow\frac{U_{n+1}}{U_n}=\frac{2^n(n+1)^2}{2^{n+1}n^2}=\frac12 \left( \frac{n+1}{n} \right)^2 <\frac34\Rightarrow \\[10ex] \Rightarrow\left( \frac{n+1}{n} \right)^2<\frac32\Rightarrow n\in(2-\sqrt6;0)\bigcup(2+\sqrt6;+\infty) \\[10ex] \forall n\in(2-\sqrt6;0)\bigcup(2+\sqrt6;+\infty)\exists n'<n,n\in R $$