# Comparison test for sequences?

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

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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