limsup and liminf of $a_n=\frac{n}{10^{\lceil log_{10}n \rceil} } $

Question

For each natural number $n\ge 1$, let $a_n=\frac{n}{10^{\lceil \log_{10}n \rceil} } $ , where $\lceil x \rceil=$ smallest integer greater than or equal to $x$. Which of the following statements are true?

$(a)$ liminf $a_n=0$

$(b)$ liminf $a_n $ does not exist

$(c)$ liminf $a_n=0.15$

$(d)$ limsup $a_n=1$

My partial solution :

For $n=10^k, k\in \mathbb{N} $, we have $a_n=1$ , so limsup $a_n\ge 1$

Again $10^{\lceil \log_{10}n \rceil} \ge 10^{\log_{10}n}=n$ and so $a_n \le 1$ which implies limsup $a_n \le 1$ . Thus limsup $a_n=1$.

Now if, $10^k \lt n \lt 10^{k+1} $ , for $ k\in \mathbb{N}\cup \{0\} $ , then $a_n\gt \frac{10^k}{10^{k+1}}=\frac1{10}$, so liminf $a_n \ge 0.1$ .This eliminates $(a)$

I need help regarding $(b)$ and $(c)$ .A hint would be sufficient. Thanks and regards.

Answer 1

For (b): a liminf always exists. (If you mean the interpretation used in the comment above, i.e. exists and is a real number, then note $a_n \ge 0$, so $\liminf a_n \ge 0$)

For (c): consider the subsequence $a_{n_k}$ where $n_k = 10^k+1$: $$a_{n_k} = \frac{10^{k}+1}{10^{k+1}} \to 0.1$$ So in particular $\liminf_{n\to\infty} a_n \le 0.1$. To see this, just slowly expand the definition of a liminf. Let $\epsilon>0$ be arbitrary. Then for any $n$ sufficiently large,

$$ \inf_{m:m\ge n} a_n \le \inf\{ a_{n_k} : n_k>n\}<0.1+\epsilon$$ so that $$\liminf_{n\to\infty} a_n = \lim_{n\to\infty} \inf_{m:m\ge n} a_n \le 0.1+\epsilon $$ Then the arbitrariness of $\epsilon>0$ gives the conclusion.

Answer 2

For (b) - the $\liminf$ of a bounded sequence always exists.

For (c): As Calvin Khor points out looking at points of the form $10^k + 1$ gives a subsequence with this limit. I just want to say a word about how you could come up with that from scratch. You already proved $\liminf a_n \ge \frac1{10}$, and you did by seeing that when $10^k<n<10^{k+1}$, the sequence becomes $$ \frac{n}{10^{k+1}} > \frac{10^k}{10^{k+1}} $$ If you want to get close to this lower bound, you simply need to pick a subsequence of the $n$ that is very close to, but just a little bigger than, $10^k$. For example $10^k+1$ works. Or $10^k+16$, $10^k+k$, $10^k + k^2$ etc.