# How to show $\lim\limits_{n\to\infty} \frac{a_n}{n}=2 \Rightarrow \lim\limits_{n\to\infty}(a_n-n)=\infty$?

I find it hard to answer the question below. I just don't know how to use the fact that $$\lim_{n\to\infty} \frac{a_n}{n}=2.$$ Maybe with limit arithmetic?

Let $(a_n)$ be a sequence, where $$\lim_{n\to\infty} \frac{a_n}{n}=2$$

Is it correct that $$\lim_{n\to\infty}(a_n-n)=\infty$$

I think it is correct since from limit arithmetic I can get to the conclusion that $$\lim_{n\to\infty}a_n=2\lim_{n\to\infty}n$$

But I just can't prove it.

Thanks.

• what is $a_n$?? – Mastrem Sep 11 '15 at 11:07
• Just a sequence, nothing I know about besides that fact. – Alan Sep 11 '15 at 11:08
• Assume that the sequence $(a_n-n)_{n \in \mathbb{N}}$ is bounded and see what you can conclude for $\lim_{n \to \infty} \frac{a_n}{n}$. – G. Bach Sep 11 '15 at 11:11
• $\frac{a_n}{n} \to 2 \implies a_n \to 2n \implies a_n - n \to n$. – user265328 Sep 11 '15 at 11:13

Hint: Prove that $\dfrac{a_n}{n} > 1.5$ for all $n$ sufficiently large.

Solution:

Since $\dfrac{a_n}{n} \to 2$, taking $\varepsilon=0.5$, we get that $\dfrac{a_n}{n} > 1.5$ for all $n$ sufficiently large. This implies that $a_n-n> 0.5 n$ for all $n$ sufficiently large and so $a_n -n \to \infty$.

$$\lim_{n\to\infty} \frac{a_n}{n}=2$$ then

$$\lim_{n\to\infty} \frac{a_n}{n} - \lim_{n\to\infty} \frac{n}{n}= 1$$

As your function is $${a_n - n}$$ the limit is

$$(\lim_{n\to\infty} \frac{a_n}{n} - \lim_{n\to\infty} \frac{n}{n}) \cdot \lim_{n\to\infty} n = 1 \cdot \lim_{n\to\infty} n = \infty$$

for $\forall n>N_{\epsilon}: \mid(\frac{a_n}{n}-2)\mid<\epsilon$ which equals to $\mid\frac{a_n-n}{n}-1\mid<\epsilon$, hence $1-\epsilon\leq\frac{a_n-n}{n}\leq1+\epsilon$. As $n>0$ we find $n-n\times\epsilon\leq{a_n-n}\leq n+n\times\epsilon$. A posteriori, choose now $\epsilon<\frac{1}{n^3}$ you will have the desired result as $n\times\epsilon<\frac{1}{n^2}<\epsilon_0$ if $n>\sqrt{\epsilon_0^{-1}}=N_{\epsilon_0}$. Here you are.

If $\lim\limits_{n\to\infty} \frac{a_n}n=2$ then $$\lim\limits_{n\to\infty} \frac{a_n-n}n=1$$ since $\frac{a_n-n}n = \frac{a_n}n-1$.

From this you get $$\lim\limits_{n\to\infty} (a_n-n) = \lim\limits_{n\to\infty} n\cdot \frac{a_n-n}n = \infty\cdot1=\infty.$$

First note that $$\lim\limits_{n\to\infty} \frac{a_n}{n}=2$$ Implies that $$\lim\limits_{n\to\infty} \frac{a_n}{2n}=1$$ Which means that $a_n$ and $2n$ are asymptotically equivalent. Therefore $$\lim\limits_{n\to\infty} \left(a_n-n\right)=\lim\limits_{n\to\infty} \left(2n-n\right)$$ $$=\lim\limits_{n\to\infty} n=\infty$$

$$\because \lim_{n\to\infty}\frac{a_n}{n}-1=\lim_{n\to\infty}\frac{a_n-n}{n}=1$$ $$\therefore a_n-n=O(n) \rightarrow \infty (n\to\infty)$$