Find a sequence such that $\liminf a_n^{1/n}=1/4,\ \limsup a_n^{1/n}=1/3$

Question

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.

Answer 1

This is not complete, but is probably enough to be getting on with (and maybe someone else will fill in the gaps).

First, to avoid writing lots of irritating fractions, I will indicate how to find an increasing sequence $\{a_n\}$ of positive reals such that

$$\liminf a_n^{1/n} = 3 \text{ and } \limsup a_n^{1/n} = 4,$$ since we can then get a solution for the original question by using the sequence $\{\frac{1}{a_n}\}$.

Note that putting $a_n = 3^n$ would give a sequence with $\liminf a_n^{1/n} = 3$, and putting $a_n = 4^n$ would give a sequence with $\limsup a_n^{1/n} = 4$, so we need to "interleave" these two sequences (and fill in some gaps) to satisfy all requirements.

Choose $a_1$ to be $3$, then $a_2$ to be $4^2$, then look for the next $k$ such that $3^k>a_2$, then then next $k$ such that $4^k>a_3$ etc, - not worrying for the moment about the missing terms:

n           3^n         4^n
1           3           
2                       16
3           27          
4                       256
5                       ??
6           729         
7                       16384
8                       ??
9           19683           
10                      1048576
11                      ??
12                      ??
13          1594323         
14                      268435456
15                      ??
16                      ??
17                      ??
18          387420489           
19                      274877906944

Now we fill in the gaps by simply increasing the previous term $4^n$ by 1 each time, and "it is easily seen that" we get:

n           3^n         4^n         a_n^1/n
1           3                       3
2                       16          4
3           27                      3
4                       256         4
5                       257         3.034
6           729                     3
7                       16384           4
8                       16385           3.364
9           19683                       3
10                      1048576         4
11                      1048577         3.526
12                      1048578         3.175
13          1594323                     3
14                      268435456       4
15                      268435457       3.647
16                      268435458       3.364
17                      268435459       3.132
18          387420489                   3
19                      274877906944        4

I would be grateful for anyone who can show me how to write this out in a sensible way!

Answer 2

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 $\ell$.

Answer 3

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?

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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.

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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 $A_n$, $B_n$, $C_n$.