Prove a sequence is cauchy if it is based on a cauchy sequence

Question

Let ${a_n}$ be a Cauchy sequence, with $a_n ≥ a > 0$. Working directly from the definitions, show that the sequence ${b_n}$ defined by $b_n = 2 + (a_n)^4$ is Cauchy.

Answer 1

Well, seeing as this is tagged under real-analysis, I'll assume $a_n \in \mathbb{R}$.

Hint: A sequence in $\mathbb{R}$ is Cauchy $\iff$ it converges, and the function $f(x) = x^4$ is continuous. Do you see where to go from here?

Answer 2

Hint: Cauchy sequence is bounded. Let $|a_n|<M$ then $|b_n-b_m|=|(a_n+a_m)(a_n^2+a_m^2)|\cdot |a_n-a_m|<4M^3|a_n-a_m|\ldots $

Inequality $a_n>a>0$ was not required.

Answer 3

$a_n$ is Cauchy $\Rightarrow$ $a_n$ is bounded so we have 0$\lt$a$\le$$a_n$$\le$M , M$\in$$\mathbb{R}$. Let $\varepsilon$$\in$$\mathbb{R}$, $\varepsilon$$\gt$0 be arbitrary , now for $\alpha$=$\frac{\varepsilon}{4M^3}$ by definintion of cauchy sequence: $\exists$ N$\in$$\mathbb{N}$, $\forall$ m,n$\in$$\mathbb{N}$ , (m,n$\gt$N)$\Rightarrow$|$a_n$-$a_m$|$\lt$$\alpha$. So now we have : $\forall$n,m$\in$$\mathbb{N}$ ,(n,m$\gt$N)$\Rightarrow$|$b_n$-$b_m$|=|$(a_n)^4$-$(a_m)^4$|=|$a_n$-$a_m$||$a_n$+$a_m$||$(a_n)^2$+$(a_m)^2$|$\lt$$\alpha$$4M^3$=$\varepsilon$. So we proved that: $\forall$$\varepsilon$$\gt$0, $\exists$N$\in$$\mathbb{N}$,$\forall$n, m$\in$$\mathbb{N}$, (n,m$\gt$N)$\Rightarrow$|$b_n$-$b_m$|$\lt$$\varepsilon$.