Prove formally that if $\lim_{n\to+\infty} a_n = +\infty$ then $\lim_{n\to+\infty}b_n = +\infty.$ Let $\left(a_n\right)_{n\in\mathbb{N}}$ and $\left(b_n\right)_{n\in\mathbb{N}}$ be two real sequences. Suppose that there exists $N_0\in\mathbb{N}$ such that $a_n \leq b_n$ for all $n > N_0$. Prove that if $\lim_{n\to+\infty} a_n = +\infty$ then $\lim_{n\to+\infty}b_n = +\infty.$
Solution 
$\lim_{n\to+\infty} a_n = +\infty$ and the sequence diverges. Since $a_n < b_n$, $b_n$ should also diverge and we can have $\lim_{n\to+\infty}b_n = +\infty.$
 A: By definition $\lim_{n\to+\infty} a_n = +\infty$ means that for any number A (no matter how large it is), there exists a positive integer number $n_1$ such that: for all $n > n_1$ we have: $a_n > A$. OK, let's pick any number A, and then find the corresponding number $n_1$. OK, now for all $n > n_1$ we have: $a_n > A$. Then for all $n > max(N_0, n_1)$ we have: $b_n >= a_n > A$. Let's denote $max(N_0, n_1) = n_2$. So for all $n>n_2$ we have: $b_n > A$. But then by the definition, it means that we also have: $\lim_{n\to+\infty} b_n = +\infty$. 
A: $$\lim _{ n\rightarrow \infty  }{ { a }_{ n }=+\infty \quad \Rightarrow \forall E>0,\quad \exists N\left( E \right) \mathbb{N} :\forall n\ge N\quad \Rightarrow \quad { a }_{ n }\ge E }  $$
$$\exists { N }_{ 0 }\epsilon \mathbb{N},  { a }_{ n }\le { b }_{ n }\quad \Rightarrow max\left( N\left( E \right) ,{ N }_{ 0 } \right) =N^{ \prime  }\Rightarrow \quad \forall n\ge N^{ \prime  }\quad { b }_{ n }\ge { a }_{ n }\ge E\quad$$

$$ \Rightarrow \lim _{ n\rightarrow \infty  }{ { { b }_{ n }=+\infty  } } $$

