Unbounded sequence and its reciprocal. Let $a_n$ and $b_n$ be 2 positive and increasing sequence. Then is it true that either the sequence  $a_n/b_n $ or sequence $ b_n/a_n $ is bounded ?
At first i thought, if $a_n/b_n $ is unbounded, then $b_n/a_n $ is bounded and vice versa. This is very intuitive to me since it's like saying: if $a_n$ grows faster than $b_n$ then $b_n $ grows slower then $ a_n$, then $b_n/a_n$ should be bounded given that they are all positive.      
But then, i think the above is wrong, $a_n/b_n$ is unbounded then the limit law $lim \frac {a_n} {b_n}\times lim \frac{b_n}{a_n} $ doesn't apply and $b_n/a_n$ can also be unbounded.
Can someone give me an example that both $a_n/b_n$ and $b_n/a_n$ are unbounded and explain why they can both grow unbounded ? 
 A: Let $$a_n=\begin{cases}
(2n)^{2n},&\text{if }n\text{ is even}\\
(2n+1)^{2n+1},&\text{if }n\text{ is odd}\;,
\end{cases}$$
and let $$b_n=\begin{cases}
(2n+1)^{2n+1},&\text{if }n\text{ is even}\\
(2n)^{2n},&\text{if }n\text{ is odd}\;.
\end{cases}$$
It's easy to check that both sequences are increasing, and since
$$\frac{(2n+1)^{2n+1}}{(2n)^{2n}}=(2n+1)\left(1+\frac1{2n}\right)^{2n}>2n+1\;,$$
$\dfrac{a_n}{b_n}>2n+1$ when $n$ is even, and $\dfrac{b_n}{a_n}>2n+1$ when $n$ is odd, so both ratio sequences are unbounded.
A: The answer is yes, the example can be constructed as follows:
Take $c_{2n-1}=n$, $c_{2n}=1/n$, and let $a_n=(n!)^2\cdot c_n$, $b_n=(n!)^2$, then we have $a_{2n-1}/b_{2n-1}=n$, and $a_{2n}/b_{2n}=1/n$. Thus both $a_n/b_n$ and $b_n/a_n$ diverge.
We may check that $a_{2n+1}/a_{2n}= (2n+1)^2\cdot n(n+1)>1$, $a_{2n}/a_{2n-1}=(2n)^2/n^2=4>1$, so $a_n$ is increasing, and $b_n=(n!)^2$ is also increasing.
A: Hmm how about this: $a_{4k} = b_{4k} = (2k)!$, $a_{4k+1} = a_{4k}+1, a_{4k+2} = a_{4k+1}k, a_{4k+3}=a_{4k+2}k$, $b_{4k+1} = b_{4k}k, b_{4k+2} = b_{4k+1} + 1, b_{4k+3}=b_{4k+2}+1$. Here $a_n/b_n$ goes to 1 for the subsequences with n = 4k and n = 4k+2, but it is unbounded for the subsequence with n = 4k+3, while its inverse is unbounded for the sbsequence with n = 4k+1.
