Calculation of Lower Box and Box Dimension I am new to this site, so sorry if the question is stupid. I am learning fractals and my teacher gave me the following exercise.

Let $$E=\{0,0\}\cup\left\{\bigcup_{n=1}^\infty (x,1/\sqrt{n}\,):0\leq x\leq 1/\sqrt{n}\right\}$$ Find $\dim_\text{lower box}(E)$ and $\dim_\text{box}(L\cap E)$ for all lines $L$ with non-zero gradient and which do not pass through $(0,0)$.

I have done some of the work but I do not know if it is correct.

$E$ can be covered by $n(n+1)/2$ boxes of side $1/\sqrt{n}$. But then this gives $$\dim_\text{lower box}(E)\leq\lim_{n\to\infty}\frac{\log(n)+\log(n+1)-\log(2)}{\frac{1}{2}\log(n)}=4$$

I am certain this is not correct. Could someone possibly please help me? Thank you very much.
Ok, now I am told that I am supposed only to show $\dim_{lower\ box}\geq\frac{4}{3}$. Does anyone know how I do this? Thanks you.

Since it is probably a homework problem, I'll just get you started.
Suppose the width of the box is $\epsilon$.  For each line segment $(0,1/\sqrt n)\times\{1/\sqrt n\}$ you need at least $1/(\sqrt n \epsilon)$ boxes, and these boxes won't cover more than one line segment if $\frac1{\sqrt n}-\frac1{\sqrt{n+1}} > \epsilon$, which gives $n \lessapprox \epsilon^{-2/3}$, since $\frac1{\sqrt n}-\frac1{\sqrt{n+1}} \approx \frac 1{n^{3/2}}$.
The number of boxes this gives can be written as a sum, and this sum can be approximated by an integral.
