I have to find ball-dimension of space $\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}$, which is $\limsup_{\epsilon\to 0}\frac{\log\beta(\epsilon)}{|\log\epsilon|}$.
Here, $\beta(\epsilon)$ is minimum cardinality of a covering of metric space by $\epsilon$-balls. Diameter of $\epsilon$ ball is $2\epsilon$, therefore $\beta(\epsilon)>\frac{1}{2\epsilon}$ and $\limsup_{\epsilon\to 0}\frac{\log\beta(\epsilon)}{|\log\epsilon|}\geq 1$.
How to conclude what is ball-dimension of this space? I'm also looking for textbook with similar problems or notes where I can read about ball dimension.
By joriki's answer dimension of space is $\frac{1}{2}$, is there any way to see it intuitively?
Any help is welcome. Thanks in advance.
Your conclusion is wrong because you don't need to cover the entire interval $[0,1]$, just the points $\frac1n$.
Given $\epsilon$, use $\epsilon^{-1/2}$ balls to cover the interval $[0,\epsilon^{1/2}]$, and cover the remaining uncovered $\epsilon^{-1/2}$ points with another $\epsilon^{-1/2}$ balls individually. That's $2\epsilon^{-1/2}$ balls, so the ball dimension is at most $\frac12$.
To see that it's not less than $\frac12$, note that around $n\sim\epsilon^{-1/2}$ the distance between consecutive values of $\frac1n$ is of the order of $\frac1{n^2}\sim\epsilon$, so you can't do any better than covering the numbers up to $\epsilon^{-1/2}$ at most two at a time, so you need at least $\frac12\epsilon^{-1/2}$ balls.
(I left out some of the details, which you may need to fill in if you want a thoroughly formal proof.)