# Prove that a_n converges

#### Alexmahone

##### Active member
Suppose $a_n$ is increasing, $a_n\le b_n$ for all $n$, and $b_n\to M$. Prove that $a_n$ converges.

My attempt:

It is sufficient to prove that $a_n$ is bounded above.

Given any $\epsilon>0$,

$|b_n-M|<\epsilon$ for $n\gg 1$

$-\epsilon<b_n-M<\epsilon$ for $n\gg 1$

$M-\epsilon<b_n<M+\epsilon$ for $n\gg 1$

$a_n\le b_n<M+\epsilon$ for $n\gg 1$

$a_n<M+\epsilon$ for $n\gg 1$

Taking $\epsilon=1$ (say) we get $a_n<M+1$ for $n\gg 1$.

So, $a_n$ is bounded above by $M+1$ for $n\gg 1$. Since $a_n$ is increasing, it is bounded above by $M+1$ for all $n$.

--------------------------------------------------------------------------------------------------------

Could someone please check the above proof for me?

Last edited:

#### PaulRS

##### Member
It seems good, the only thing I do not like is the fact that you write " for $n \gg 1$ " instead of saying "there exists $n_{\epsilon} \in \mathbb{N}$ such that... " or "for large enough $n$". • Alexmahone

#### ThePerfectHacker

##### Well-known member
Here is a similar way, but this idea is useful in its own right.

Theorem: If $(b_n)$ is a convergent sequence then it is bounded.

Proof: Left as exercise ---> you basically proved it above.

Corollary: If $(a_n)$ is increasing and $(b_n)$ is convergent with $a_n\leq b_n$ then $(a_n)$ is convergent.

Proof: There is $A$ such that $b_n \leq A$ by theorem but then $a_n \leq A$ as well and so .... QED

Corollary: If $(a_n)$ is decreasing and $(b_n)$ is convergent with $a_n \geq b_n$ then $(b_n)$ is convergent.

Proof: There is a $B$ such that $b_n \geq B$ by theorem but then $a_n \geq B$ as well and so .... QED