TITLE: Find the limit infimum of $\frac{n}{10^{[\log_{10}n]}}$
QUESTION [0 upvotes]: Q. Consider the sequence $\{a_n\}_{n \geq 1}$, where $a_n=\frac{n}{10^{[\log_{10}n]}},~~[x]$ denotes the smallest integer greater than or equal to $x$. Find $\liminf_{n\rightarrow \infty}a_n$ and $\limsup_{n\rightarrow \infty}a_n$
Since $\log_{10}{n_k} =k~\in \mathbb N$ if $n_k=10^k$, we have
$$\frac{n_k}{10^{[\log_{10}n_k]}}=1$$  and thereby $$\limsup_{n\rightarrow \infty}a_n=1$$how to find $\liminf_{n\rightarrow \infty}a_n$?

REPLY [1 votes]: We have $\lceil \log_{10}n \rceil - 1 < \log_{10}n \le \lceil \log_{10}n \rceil$ just from how rounding up works. Therefore
$$
    10^{\lceil \log_{10}n \rceil -1} < 10^{\log_{10} n} \le 10^{\lceil \log_{10}n \rceil } \implies \frac1{10} \cdot 10^{\lceil \log_{10}n \rceil} < n \le 10^{\lceil \log_{10}n \rceil }.
$$
Dividing through by $10^{\lceil \log_{10}n \rceil}$ gives us
$$
   \frac1{10} < \frac{n}{10^{\lceil \log_{10}n \rceil}} \le 1. 
$$
We can get closer and closer to $\frac1{10}$ every time $n = 10^k+1$, for instance: in that case, $10^{\lceil \log_{10}n \rceil} = 10^{k+1}$, and we get $\frac{n}{10^{\lceil \log_{10}n \rceil}} = \frac1{10} + \frac1{10^{k+1}}$. So that is in fact the $\liminf$.
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