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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

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

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 $n$.

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Could someone please check the above proof for me?

**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 $n$.

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Could someone please check the above proof for me?

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