Can this sum ever converge? If I have a strictly increasing sequence of positive integers, $n_1<n_2<\cdots$, can the following sum converge?
$$ \sum_{i=1}^\infty \frac{1}{n_i} (n_{i+1}-n_{i}) $$
I suspect (and would like to prove) that it always diverges. Haven't made much progress so far, though. 
On a related note, is there any characterization of which subsequences of $1/n$ have a convergent sum? 
 A: To expand on my comment (oops, and even more on Achille Hui's comment which appeared when I was typing):
$$
\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_i}^{n_{i+1}}\frac{1}{t}dt\;\Rightarrow \; \sum_{i=1}^K\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_1}^{n_{K+1}}\frac{1}{t}dt=\log\left(\frac{n_{K+1}}{n_1}\right)\rightarrow +\infty.
$$
Of course, $n_K$ tends to $+\infty$ as $n_k\geq K$ for every $K$ by induction.
For your other question, a partial answer can be given with the help of the notion of natural/asymptotic density. If $\liminf \frac{k}{n_k}>0$, then there exists $m>0$ and $K$ such that $\frac{k}{n_k}\geq m$ for every $k\geq K$, whence $\sum_{k\geq K}\frac{1}{n_k}\geq m\sum_{k\geq K}\frac{1}{k}=+\infty$. That is, if the set $\{n_k\;;\;k\geq 1\}$ has positive asymptotic lower density, then the series $\sum_{k\geq 1}\frac{1}{n_k}$ diverges. But the converse is false. For instance, the set of prime numbers $\mathcal P$ has null asymptotic density and yet $\sum_{p\in\mathcal{P}}\frac{1}{p}=+\infty$. 
