Apparent contradiction for probability density functions? Consider a probability density function $\it{pdf}$, $f\left(x\right)$, which can be expanded as:
$$ f\left(x\right) = \sum_{k=1}^{\infty} \alpha_k \delta\left(x-x_k\right)$$
It is easy to verify, by integrating both sides that, $\int_{-\infty}^{\infty}f\left(x\right) = 1$, implies $\sum_{k=1}^{\infty}\alpha_k=1$. Also the orthogonality of dirac delta functions centered on distinct points $x_k$ and $x_m$ leads to $\alpha_k=f\left(x_k\right)$. Here's the problem, the only bound on a $\it{pdf}$ is non-negativity, so since $f\left(x\right) \in \left[0,\infty\right)$, we should have $\alpha_k \in \left[0,\infty\right)$. If $\alpha_k$ has no upper bound, how can they sum to $1$?
 A: Your implication $\sum_{k=1}^{\infty}\alpha_k=1$ is correct. The problem is your assumption $\alpha_k=f(x_k)$. This is not the case, since the delta function is defined by its integral; it is a distribution and does not have a function value in the traditional sense. So evaluating $\delta(x-x_k)$ at $x=x_k$ is meaningless.
EDIT: This is a reaction to your example in the comment below.
$$f(x_m)=\int_{-\infty}^{\infty}f(x)\delta(x-x_m)\;dx =
\int_{-\infty}^{\infty}\sum_{k=1}^{\infty}\alpha_k\delta(x-x_k)\delta(x-x_m)\;dx$$
Assuming that we can interchange integration and summation, we get
$$f(x_m) = \sum_{k=1}^{\infty}\alpha_k\int_{-\infty}^{\infty}\delta(x-x_k)\delta(x-x_m)\;dx$$
Now we evaluate the integral in a way that is common in physics and engineering; I believe mathematicians usually frown upon such manipulations (please correct me if I'm mistaken):
$$f(x_m) = \sum_{k=1}^{\infty}\alpha_k\delta(x_m-x_k)\tag{1}$$
which is not the same expression that you derived in your comment. Note that (1) is exactly what you get when you evaluate $f(x)$ at $x=x_m$ in your original expression for $f(x)$. So $f(x_m)\neq \alpha_m$, because here, $\delta(x)$ is not the Kronecker delta, but the delta function.
A: Matt has correctly pointed out one problem with your question, namely that it makes no sense to evaluate a distribution (as opposed to a function) at a point.  There is, however, an additional problem with the question.  You wrote "the only bound on a pdf is nonnegativity".  That's the only bound that refers to values of a pdf at just one point at a time (when the pdf is a function, not merely a distribution, so that it has values at points), but the requirement that the integral be $1$ amounts to an additional, global bound.  You can't just have  arbitrarily large values everywhere and expect the integral to be 1.  A pdf can have an arbitrarily large value at one point (or at a few points) but it will have to have small values elsewhere to make the integral $1$.  
A: But it does!! Since 
$$\int_{-\infty}^{\infty} f(x)dx =1 \Rightarrow \sum_{k=1}^{\infty} \alpha_k = 1$$
Now we have $\alpha_k = f(x_k) \geq 0$, and since $\sum_{k=1}^{\infty} \alpha_k$ is a convergent series, we also have $\alpha_k \to 0$. Moreover 
$$\alpha_n \leq \sum_{k=1}^{\infty} \alpha_k = 1$$
Thus $0 \leq \alpha_n \leq 1$ for all n.
By chance I have misunderstood your question this might help:
$$\alpha_k \delta{\{x-x_k\}} \leq \sum_{k=1}^{\infty} \alpha_k\delta{\{x-x_k\}} = f(x)$$
Integrate both sides to get (Noting all terms are nonnegative) that
$$\alpha_k \leq 1$$
