Find a nonnegative sequence of real numbers satisfying certain conditions Let $a_n$ be a nonnegative and nondecreasing sequence and $\lim a_n = \infty$, i.e. $a_n \uparrow \infty$ (1), also we know that $\lim \frac{a_{n+1}}{a_n} = 1$ (2). Find such sequence which also satisfies $\lim a_{2n}/a_n = \infty$. (3)
I tried hard, but failed to find such sequence. Then I think of trying to prove that if $\lim\frac{a_{n+1}}{a_{n}}=1$ (2), then $\lim\frac{a_{2n}}{a_n}=c$ for some constant $c\geq 1$ (4). But unfortunately I fail to prove it.
So to sum up my question: Find a nonnegative sequence satisfying (1),(2),(3). Or prove that if a nonnegative sequence satisfies (1), (2), we must have (4). 
Really appreciate for any comment.
 A: We can define $a_n$ recursively by taking $a_1=1$ and $a_{n+1}=a_n\times n^{1/n}$
Clearly we have $\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim\limits_{n\to \infty}n^{1/n}=1$.
On the other hand $\lim\limits_{n\to \infty}\frac{a_{2n}}{a_n}=\lim\limits_{n\to \infty}(n+1)^{1/(n+1)}\dots (2n)^{1/2n}\geq \lim\limits_{n\to \infty}((n)^{1/2n})^n=\lim\limits_{n\to\infty}\sqrt n =\infty$
A: Sequence could be set to $$a_n = \left(1+\frac{1}{\log(n)}\right)a_{n_1}$$
You can prove that 
$$\frac{a_{2n}}{a_n} = \prod_{i=n}^{2n}\left(1+\frac{1}{\log(i)}\right)\rightarrow \infty$$
while $\frac{a_{n+1}}{a_n} \rightarrow 1$.
A: 
What about $(a_n)_n$ defined by $a_1=1$ and
  $$
a_{n+1} = \left(1+\frac{1}{\sqrt{n}}\right)a_n
$$
  for $n\geq 1$?

We have


*

*$(a_n)_n$ is increasing, and $a_n = \prod_{k=1}^n\left(1+\frac{1}{\sqrt{k}}\right)$, from which
$$
\ln a_n = \sum_{k=1}^n \ln\left(1+\frac{1}{\sqrt{k}}\right) \operatorname*{\sim}_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{k}} \xrightarrow[n\to\infty]{} \infty
$$
so (1) holds.

*We have $$\frac{a_{n+1}}{a_n} = 1+\frac{1}{\sqrt{n+1} }\xrightarrow[n\to\infty]{} 1$$
so (2) holds.

*We have $\frac{a_{2n}}{a_n} = \prod_{k=n+1}^{2n}\left(1+\frac{1}{\sqrt{k}}\right)$ from which
$$\begin{align}
\ln \frac{a_{2n}}{a_n} &= \sum_{k=n+1}^{2n} \ln\left(1+\frac{1}{\sqrt{k}}\right)
= \sum_{k=1}^{n} \ln\left(1+\frac{1}{\sqrt{n+k}}\right) 
\\&\geq 
n \ln\left(1+\frac{1}{\sqrt{2n}}\right) 
\operatorname*{\sim}_{n\to\infty} \sqrt{\frac{n}{2}}\xrightarrow[n\to\infty]{} \infty
\end{align}$$
so (3) holds.
