How can expected value be infinite? My book as well as Wikipedia 
gives this definition of expected value:

$\mathbb E(X)=\sum _x xf(x).$ But, $\mathbb E(X)$ is said to exist if and only if that equation is absolute convergent. 

But, I see that many places do not follw this def., e.g. here $${\mathbb E} X = \sum_{n=1}^\infty 2^{-n} \cdot 2^n = \sum_{n=1}^\infty 1 = \infty,$$ though we can see that absolute convergence test is failed.
So, can expected value be infinity or negative infinity?
 A: The expected value of your example does not exist. However, since the series that would define the expected value as a generalized limit of $\infty$, we sometimes (sloppily) say that the expected value is $\infty$. As long as you know what you are talking about, that should not be too big of a problem.
A: Yes. It can be. Here is an example that I faced in one of my works.
Assume $X$ to be an Exponential distribution ($f_X(x)=e^{-x}$) and $Y=\frac{1}{X}$. For this case, $\mathbb{E}(Y)=\infty$.
Indeed, writing the expectation as integral:
$$
   \mathbb{E}(Y) = \int_0^\infty \frac{1}{x} \mathrm{e}^{-x} \mathrm{d} x
$$
you see that the integral diverges at the lower bound. Thus, while it is natural to expect $\mathbb{E}\left(Y=\frac{1}{X}\right) > 0$, the expectation  is infinite.
A: I just answered a similar question on SE stats, CrossValidated.  Rather than copy that answer over here, I give the link:
https://stats.stackexchange.com/questions/94402/what-is-the-difference-between-finite-and-infinite-variance
That answer is much longer and more detailed than the ones above here!
