Uniqueness of moments for probability distributions with infinite moments. I was taught the collection of a distribution's moments uniquely defined the distribution. Recently, I have been studying Pareto distributions, which have infinite means for shape parameters less than 1. It was also my understanding that if a moment of rank $\rho$ does not converge, all moments of rank greater than $\rho$ will also not converge. As such, for a Pareto distribution with shape $\alpha$ less than 1, its moments are:
$moment = \left\{
     \begin{array}{lr}
       \infty & : order \ge 1\\
       1 &      : order = 0
     \end{array}
   \right.\\$
The moments are identical for multiple different Pareto distributions, violating my initial statement, a contradiction. 
Could anyone help point out where my lack of understanding lies?
Edit: fixed some notation
 A: Getting the right distribution from moments is a tricky business. First off, the sequence of moments has to be valid. You can see a technical condition involving Hankel matrices in the link. Basically, it says you can't just write down an arbitrary sequence of numbers and claim they are moments of some distribution. In your case, the moments actually come from a distribution, so you don't have to worry about this.
In the case that all moments exist (are finite), then there are two subcases. If your distribution is supported on a finite interval $[a,b]$ then there is a unique measure that corresponds to them. If the interval is infinite, then the measure is not necessarily unique. You need an extra condition, something like Carleman's condition to guarantee uniqueness. 
Finally, there's your case, which is a truncated moment problem: you only have existence of the first $k$ moments. This problem has even less uniqueness. This is because you can cook up plenty of distributions, such as power laws that look like $d/(1+a(x-b)^c)$ which can sometimes match all your moments. In your example, take $1/(1+|x-1|^3),$ which has mean 1 but no second moment. 
A: There is a one-to-one correspondence between distributions that
have moment generating functions (MGFs) and their MGFs. But not
every distribution has an MGF (and some have MGFs that are
not of practical use because of analytic difficulties). 
If you
have a particular MGF, you can be sure only one distribution matches
that MGF.
By taking the $k$th derivative $m_X^{[k]}(t)$ of the MGF of $X$ and setting $t = 0$
you get $E(X^k) = m_X^{[k]}(0).$ Conversely, if you know
all of the moments $E(X), E(X^2), \dots,$ you can use them
to write a Taylor series that is equal to $m_X(t).$ (See Wikipedia
on 'moment generating function' in the Definition section.)
It is in that sense that some texts say that a distribution
is uniquely specified by its moments. However, in practice,
I do not think this is often a useful way to identify a distribution.
Note: Outside of statistics and probability MGFs are called Laplace transforms. Characteristic functions (Fourier transforms) exist
for all distributions, and they can also be used to find moments,
but are somewhat more advanced mathematically than MGFs. I'm guessing
it is the connections among MGFs, distributions, and moments, as explained above, that prompted your question.
