Sum of the inverse of degrees in graph Suppose $G$ is an infinite and locally finite graph. Are there easy conditions (and non-trivial!) conditions on $G$ guaranteeing that
$$\sum_{v \in V(G)} \frac{1}{\deg v} < \infty \;\;?$$
Here $\deg v$ is just the valence of the vertex $v$.
Moreover, if this series converges, is there any interpretation of the resulting number?
 A: If a graph has $V$ vertices, then the minimum value of your series takes place when you have a complete graph unless you want it to be possible for multiple edges to connect two nodes. Even still, if the number of edges you allow to connect two vertices is a constant, the minimum value still takes place at this modified version of a complete graph. The sum of your series is therefore $\frac{V}{k(V-1)}$, where $k$ is the maximum number of edges between any two nodes. If you allow loops from a node to itself, then the sum equals $\frac1k$. Now, for an infinite graph, we take $\lim_{V\to\infty}\frac{V}{k(V-1)}=\frac1k$. We can extend this by introducing a new variable, $c$, which indicates an "average" (assuming that the edges are distributed evenly, more on that later) of what fraction of the other nodes a node connects to. Explicitly, $c=\frac{2E}{V^2}$, which means your finite sum equals $\frac V{ck(V-1)}=\frac{V^3}{2kE(V-1)}$. This will only converge if $E=\Theta(V^2)$, but since you plan on having an infinite number of edges, as long as $c>0$, you'll be fine. This leaves us with your sum being $\frac1{ck}$. Note that lower values of $c$ yield a higher value for your sum and higher values of $k$ yield a lower value.
This, however, assumes that the edges are spread out equally among all the nodes. As a simple example, imagine a finite graph with ten nodes and ten edges. For all connected graphs like this $c=\frac15$ and $k=1$. If the graph was a cycle, the sum would equal $5$. If the graph was instead this:

The sum would be $\frac{73}9=8\frac19$. This shows us that the less even the distribution of vertices on the graph is, the higher the sum. In other words, if you can express the number of edges as a function of the number of vertices, the difference between the sum and $\frac V{ck(V-1)}=\frac{V^3}{2kE(V-1)}$ will measure how unevenly spread the edges are. This will only be a problem, however, if you have an infinite number of nodes that have a finite number of neighbors.
A fun infinite graph that I can think of is this one:

which you can generate by creating adding a row with double the number of nodes in the row before it and connecting every node in every row with the node below it. The number of edges between the $n-1^{th}$ and the $n^{th}$ row is $2^n2^{n-1}=2^{(n-1)n}$, so the total number of edges of the $n^{th}$ partial graph is  $$\sum_{k=0}^n2^{k(k+1)}>2^{n^2}$$ The total number of vertices is $$\sum_{k=0}^{n+1}2^k=2^{n+2}$$
which means this converges. I have no idea how to find a partial sum for the number of edges, so I just used a pretty good lower bound. Regardless, the number of edges would be so far above the number of vertices that your sum would go to zero if the edges were distributed evenly. This graph, however, does not distribute the edges evenly. A node in row $k$ touches exactly $2^{k-1}+2^{k+1}$ other nodes, and there are $2^k$ nodes in each row. Your sum would be $$\frac12+\sum_{k=1}^\infty\frac{2^k}{2^{k-1}+2^{k+1}}=\frac12+\sum_{k=1}^\infty\frac{2^k}{2^{k-1}(1+2^2)}=\frac12+\sum_{k=1}^\infty\frac25=\infty$$
This should make sense because a node in any finite row touches a finite number of nodes. With the way these nodes are organized, none of them touch a non-zero fraction of the nodes.
Finally, let's look at an infinite graph with an uneven distribution of edges. Let's say that, in this infinite graph, half the nodes touch half the nodes and half the nodes touch a third of the nodes. Other than that, all edges are spread out as equally as possible. This is what a finite version of that graph would look like

The fraction of nodes an average node touches is $$\frac12\cdot\frac12+\frac12\cdot\frac13=\frac5{12}$$
and so, a perfectly even graph would have a sum of $\frac{12}5$. This graph, however, would have a sum of $$\frac12\cdot\frac1{\frac13}+\frac12\cdot\frac1{\frac12}=\frac12\cdot(3+2)=\frac52$$
The difference of $\frac1{10}$ indicates the unevenness of the graph. I feel like you can make a continuous version of what I did in the last line of math, but I don't see how right now.
Summary: For your sum to converge, you must have only a finite number of nodes with finite degree and every other node must touch at least a nonzero fraction of all other nodes. The interpretation of the sum given some relationship between the number of vertices and the number of edges is the unevenness of the graph when compared to a graph with all edges spread out evenly among all nodes.
Quick P.S. This only applies for connected graphs as one unconnected node leads to a diverging sum.
