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Let $a_n, b_n$ such that for sufficiently large $n$: $ a_n \le b_n$.

Can we deduce that:

  1. $\lim_{n\to\infty}a_n = \infty \implies \lim_{n\to\infty}b_n = \infty$
  2. $\lim_{n\to\infty}b_n = L \implies \lim_{n\to\infty}a_n = L$, where $L < \infty$.

By the way,
I am aware of the squeeze theorem though I wondered if "one-sided-squeeze" is valid like mentioned above.

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I think there may be a typo or two in your second assertion. – msteve Jul 19 '14 at 16:19
up vote 6 down vote accepted

As others have stated, the implication in 1. is correct.

Unfortunately, there is only one thing you can deduce from the assumptions of 2. That is that if $\lim \limits_{n \rightarrow \infty} a_{n}$ exists, then we know it is not equal to $\infty$. But it could be $- \infty$. Furthermore, the limit might not exist at all. Here is a counterexample where $a_{n} \leq b_{n}$ and $\lim \limits_{n \rightarrow \infty} b_{n} = L$ for some $L < \infty$, but $\lim \limits_{n \rightarrow \infty} a_{n}$ does not exist:

Let $b_{n} = 2$ for all $n$, and $a_{n} = (-1)^{n}$. For all $n$, $a_{n} \leq b_{n}$, and $\lim \limits_{n \rightarrow \infty} b_{n} = 2$ (where clearly $2 < \infty$), but $\lim \limits_{n \rightarrow \infty} a_{n}$ does not exist.

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The second assertion is false as anorton and user46944 proved.

The first one is true. Here's a proof:

Assume that $\lim\limits_{n\to +\infty}a_n=+\infty$.

Then by the definition we get:

$\forall A>0,\,\exists N\in\mathbb{N},\,\forall n\in\mathbb{N},\, n>N\Longrightarrow a_n>A$

for sufficiently large $n$: $a_n\le b_n$

Let $N'\in\mathbb{N}$ such as $\forall n\in\mathbb{N},\,n>N'\Longrightarrow a_n\le b_n$

Let $A>0$. So $\exists N\in\mathbb{N},\,\forall n\in\mathbb{N},\, n>N\Longrightarrow a_n>A$

Let $N''=max(N,N')$. Then $N''\ge N'$ and $N''\ge N$.

So $\forall n\in\mathbb{N},\, n>N''\Longrightarrow (a_n\le b_n$ and $a_n>A)\Longrightarrow b_n>A$

We conclude that:$$\forall A>0,\,\exists N''\in\mathbb{N},\,\forall n\in\mathbb{N},\, n>N''\Longrightarrow b_n>A$$ which proves that $\lim\limits_{n\to +\infty}b_n=+\infty$.

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Your first implication is correct, but the second is not.

For a counterexample to assertion $2$, consider $a_n = 1$ and $b_n = 2$. In this case, $\lim_{n\to\infty} b_n = 2$, but $\lim_{n\to\infty}a_n = 1 \ne 2$

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