# $\lim_{n \rightarrow \infty} \frac {a_n}{b_n} = \alpha$ where $a_n$ converges to infinity but $b_n$ diverges.

I encounter the problem:

Give an example of a sequence $a_n$ that converges to infinity and $b_n$ does not diverge to positive or negative infinity and where $\lim_{n \rightarrow \infty} \frac {a_n}{b_n} = \alpha$ for some $\alpha \in \Bbb R$.

Thoughts: I have done a few problems similar to this where I have to define the sequences in piece-wise components or with a $(-1)^n$ but this particular criterion is tricky. I also realize that $a_n = \alpha b_n$ so that $a_n$ is some scalar multiple of $b_n$. Any hints much appreciated.

If you mean to say $\frac{a_n}{b_n}\to\alpha$, then $a_n=n$, $b_n=(-1)^n n^2$ should do the case, since $a_n\to\infty$ and $b_n$ does not converge. In this case, $\alpha=0$.

Interestingly, you can show that this is only possible if $\alpha=0$, i.e., you can show that if $\lim_{n\to\infty} a_n=\infty$ and $\lim_{n\to\infty}\frac{a_n}{b_n}=\alpha\neq 0$, then $\lim_{n\to\infty}{b_n}=\infty$ or $-\infty$.

Before your edit: (this is no longer relevant as you changed the question a little)

However, if you mean that $\frac{a_n}{b_n}=\alpha$ for every value of $n$, then it is impossible for $a_n$ to converge to $\infty$ and $b_n$ not converge to either $\infty$ or $-\infty$.

• I thought $\alpha$ is supposed to be independent of $n$ – Ben Grossmann Jan 26 '17 at 14:02
• Sorry I misread the question, posted my edit. – IntegrateThis Jan 26 '17 at 14:04
• I just saw another part to the question stating that $a_n$ and $b_n$ are positive, is this possible? – IntegrateThis Jan 26 '17 at 14:25
• @JamesDickens I think it's best if you ask that in another question. Better yet, try to solve it yourself first, then ask if you are still stuck. – 5xum Jan 26 '17 at 14:26
• @5xum just for good measure I chose $a_n = n$, $b_n = abs(nsin(1/n))$ – IntegrateThis Jan 26 '17 at 14:46