Find a sequence such that $\liminf a_n^{1/n}=1/4,\ \limsup a_n^{1/n}=1/3$ How to construct a sequence $\{a_n\}$ such that $1>a_n>a_{n+1}>0$ for all $n>0$ and 
$$
\liminf a_n^{1/n}=1/4,\ \limsup a_n^{1/n}=1/3.
$$ 
Thanks.
 A: Both sequences ${1\over 3^n}$ $\>(n\geq0)$ and ${1\over 4^n}$ $\>(n\geq1)$ tend to zero. By alternating between these two sequences we can produce a monotone sequence $(a_n)_{n\geq1}$ with the required properties.
Define the strictly increasing sequence $(n_k)_{k\geq1}$  as follows: Put $n_1:=1$ and for $k\geq1$ define recursively
$$n_{2k}:=\min\bigl\{j\>|\>j>n_{2k-1},\  4^j>3^{n_{2k-1}}\bigr\}, \quad
n_{2k+1}:=\min\bigl\{j\>|\> j>n_{2k},\ 3^j>4^{n_{2k}}\bigr\}\ .$$
Then let
$$a_{n_{2k-1}}:={1\over 3^{n_{2k-1}}}\ ,\quad a_{n_{2k}}:={1\over 4^{n_{2k}}}\qquad(k\geq1)\ .$$
It follows that $a_{n_{r+1}}<a_{n_r}$ for all $r\geq1$. Finally for $n_r<\ell<n_{r+1}$ choose the $a_\ell$ equidistant between $a_{n_r}$ and $a_{n_{r+1}}$.
In this way the $a_\ell$ are monotonically decreasing; and furthermore one has ${\root l \of a_\ell}={1\over 3}$ as well as ${\root l \of a_\ell}={1\over 4}$ for infinitely many $\ell$.
A: Let $b_n=\frac1{4^n}$ and $c_n=\frac1{3^n}$. It is clear if for each $n$ we have either $b_n\le a_n\le c_n$ and, moreover, $a_n=b_n$ is fulfilled for infinitely many $n$'s, then $\liminf \sqrt[n]{a_n}=\frac14$. Similarly if we have have $a_n=c_n$ infinitely often, then $\limsup \sqrt[n]{a_n}=\frac13$.
It is very simple to achieve this -- we simply divide $\mathbb N$ into two infinite sets and put $a_n=b_n$ for numbers from one set and $a_n=c_n$ for numbers from the other set.
The question is: Can we modify this somehow to get monotone sequence?

I will deal with a slightly simpler problem where I only require $a_{n+1}\le a_n$. (I do not think it should be very difficult to modify this to a strictly decreasing sequence.) 
Let us describe our sequence inductively:


*

*We start by choosing $a_n=\frac1{3^n}$ for $a\in [A_0,B_0)$, where $A_0=1$, and $B_0$ is arbitrary.

*Next we put $a_n=\frac1{4^n}$ for $a\in [B_0,C_0)$. Again $C_0$ can be chosen arbitrarily, we only want $C_0>B_0$.

*On the next interval the sequence will be constant, i.e., we put $a_n=\frac1{4^{C_0}}$ for $n\in[C_0,A_1)$. The number $N_1$ will be start of the next interval, i.e., we want to define $a_{A_1}=\frac1{3^{A_1}}$. We want to do this in such way, that our sequence will be monotone. For this we need
$$\frac1{3^{A_1}} \le \frac1{4^{C^0}}.$$
We also want that our sequence is always between $\frac1{4^n}$ and $\frac1{3^n}$. So we want
$$\frac1{4^{C^0}} \le \frac1{3^{A_1-1}}.$$
So we have to choose
$$A_1=\min\{k\in\mathbb N; \frac1{3^k} \le \frac1{4^{C^0}}\}.$$


Now we can repeat the same process inductively.

The above does not give an explicit formula for $a_n$. If you really want such a formula, you can try to play with choice of $A_n$, $B_n$, $C_n$.
