In my book it is said that Dirac function $\delta(\tau)$ can be expanded as:
$$ \delta(\tau)=(\beta \hbar)^{-1}\sum_{n \in even} e^{-i\omega_n\tau} $$
where $\omega_n=\frac{n\pi}{\beta\hbar}$, and $\tau\in(-\beta \hbar, \beta\hbar)$.
I know that Dirac function is an even function, but how can the above equation be derived?
First, although it is certainly true that an infinite sum of numbers cannot be convergent if the summands do not go to zero, for functions the corresponding criterion for pointwise convergence is by far not the only possible interpretation or sense, although it is the most elementary. So, right, the infinite sum you wrote, or any obvious variant, does not converge pointwise, but we are not necessarily obliged to care, if we have another viable-and-useful interpretation.
Another point is that, although the question declares that the argument for the (generalized) function is just a finite interval, the Fourier series naturally gives an extension to a periodic function on all of $\mathbb R$. So, if the expression makes sense, it really produces a periodic version of Dirac's $\delta$, namely, what is often called a Dirac comb.
So, in what sense does that Fourier series converge? A cautious guess would be that it converges distributionally, and this is correct, although there is some cognitive dissonance about why this should be, namely, that the Dirac comb is really the Dirac $\delta$ on the circle, and test functions on $\mathbb R$ are not immediately test functions on the circle. But if we reformulate the question so that it asks about functions on the circle, then the statement that the indicated series converges distributionally is exactly the assertion that a periodic smooth function is equal (pointwise) to its Fourier series. (Note that a test function on $\mathbb R$ need not have matching values at the endpoints of your interval, so there'll be some awkwardness there.)
In fact, that Fourier series (viewed as being a generalized function on the circle) converges in a Sobolev space with index $-{1\over 2}-\epsilon$ for every positive $\epsilon$, but maybe this is not critical to your concerns.