The density of states for free electron in conduction band In Introduction to Solid State Physics, eighth edition, by Kittel, page 141, eqs. (20,21), the density of states for electron in conduction in three dimensions is
$$D(\epsilon)\equiv \frac{dN}{d\epsilon}=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)\epsilon^{1/2}=\frac{3}{2} \frac{N}{\epsilon}.$$
The same argument could apply such that in two dimensions
$$D(\epsilon)= \frac{2}{2}\frac{N}{\epsilon},$$
and in one dimension
$$D(\epsilon)= \frac{1}{2}\frac{N}{\epsilon}$$
—i.e. with each increase of dimension, the density of states increase by $\frac{1}{2}\frac{N}{\epsilon}$.

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*Why does each dimension increase the density of states $D(\epsilon)$ by $\frac{1}{2}\frac{N}{\epsilon}$?


*What's the order of $D(\epsilon)$ in 3D, $\epsilon^{1/2}$ or $\epsilon^{-1}$?
 A: Your question can be reduced to the following:  How is the instantaneous value of a function $f(x)$ related to the average value of that function from $x_o$ to $x$?  In the case of Fermi statistics in 1, 2 and 3 dimensions we know the average value precisely by inspection.  In this formulation the average value of a function over the interval is just:
$$ \begin{equation} \overline{f(x)} = \frac{\int{f(x)dx}}{(x-x_0)} \end{equation}$$
So the ratio of the instantaneous value to the average value is:
$$ \begin{equation} \frac{f(x)}{\overline{f(x)}} = \frac{(x-x_0)f(x)}{\int_{x_0}^x {f(x)dx}} \end{equation}$$
If you work this through with the 3 functional forms of the density of states for 1,2 and 3 dimensions:
$$ \begin{equation} f(x) = (x-x_0)^{1/2}, f(x) = const, f(x) = (x-x_0)^{-1/2} \end{equation}$$
you will see that the factors of $3/2$, $1$ and $1/2$ come from the integral. Everything else cancels out so that the ratio is always just a number dependent on the shape of the function.  To see how this relates to Fermi statistics and the ideas in Kittel... keep reading.
The Fermi energy is the energy below which all states are filled at 0 K.  That means there are an equal number of states and electrons below $\begin{equation} \epsilon_F \end{equation}$ and hence you would expect the "average" density of states
up to the Fermi energy to be about $\begin{equation} \frac{ N }{\epsilon_F} \end{equation}$, one electron per orbital.  That would also be the density of states at the Fermi Energy if density of states were constant with energy... but in 3D it is not, it increases with energy as $\begin{equation} \epsilon^{1/2} \end{equation}$ so you would expect the density of states to be higher than average at the highest occupied energy, $\begin{equation} \epsilon_F \end{equation}$, so the factor in front of $ \frac{N}{\epsilon} $ should be greater than 1.  You need to go through the math above so see that it is 3/2.
The order of the density of states is $\begin{equation} \epsilon^{1/2} \end{equation}$, N is also a function of energy in 3D.
In 2D, the density of states is constant with energy.  To see this first note that energy isoquants in k-space are circles.  Valid states are discrete points in k-space.  The energy of states on a circle increases as the radius squared, $k^2$. For a change in energy $\begin{equation} \Delta \epsilon \propto  2k \Delta k\end{equation}$ the area of the circle increases as $\begin{equation} 2\pi k \Delta k \propto \Delta \epsilon \end{equation}$.  The number of states in this annular ring is proportional to this area, one unit area for each valid k-vector, so the density of states, $\begin{equation} \frac{dN }{d\epsilon} \end{equation}$ is constant.  So you would expect, in this case, for the density of states to be the average density at any energy $\begin{equation} \frac{N }{\epsilon} \end{equation}$.  In this case the numerical constant is clear, the density of states is the same at any energy so the number of electrons must exactly equal the number of states up to that energy... no math required!
The argument for 1D is similar to the argument for 3D except that in this case the density of states is decreasing as energy increases.  Constant energy isoquants in k-space in 1D are just points on a line at a distance k from the origin.  Moving one point outward from the origin increases the number of states by one and increases the energy of the state by a constant increment, $\begin{equation} \Delta \epsilon \propto 2k \Delta k\end{equation}$.
$$\begin{equation} \frac{dN }{d\epsilon} =\frac{\Delta N }{\Delta k}\frac{\Delta k }{\Delta \epsilon} \propto 1 \cdot \frac{1}{k} \propto \epsilon^{-\frac{1}{2}}\end{equation}$$
Now we would expect the density of states to be less than the average density at any energy $ \epsilon $ so the factor in front of $ \frac{N}{\epsilon} $ should be less than 1.
