Is this what Convergence/Divergence comes down to? In trying to understand why $\sum\limits_{k=1}^{\infty} \frac{1}{2^k}$ converges but $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$ doesn't, I noticed that in infinite series of the type $\sum\limits_{n=1}^{\infty} \frac{1}{k^n}$ where $k > 1$, any term is greater than the sum of any number of subsequent terms.
Whereas for example in the series $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$, it is always possible to find for any term a certain number of subsequent terms whose sum is greater than that term.
So I'm wondering, is there anything to this? Is this principle the difference between a series converging or diverging? 
 A: Convergence of series always goes back to the definition with the limit of the partial sums.  You have touched on some parts that are true.  For example, you say (paraphrasing) that if one term, say $a_{1}$ is bigger than the sum of all the rest (for example when $a_{n}=\frac{1}{2^{n}}$) then the series converges.  (Assuming all terms are positive) When this is true, you get convergence because the sum is bounded above: 
$$\sum_{n=1}^{\infty} a_{n} = a_{1}+\sum_{n=2}^{\infty}a_{n}\leq 2a_{n}<\infty.$$
The reason this implies convergence is that the sequence of partial sums $\sum_{n=1}^{N}a_{n}$ is an increasing sequence (since all the terms are non-negative) and is bounded above (by $2a_{1}<\infty$), so it converges.
One thing you should be careful of though is that this isn't true whenever $a_{n}=\frac{1}{k^{n}}$ for some $k>1$.  The series will converge, but $a_{1}\leq \sum_{n=2}^{\infty}a_{n}$ if $1<k<2$.  What you are catching here is essentially the ratio test (when subsequent terms drop in size quickly the series converges).
On the other hand, series like $\sum_{n=1}^{\infty}\frac{1}{2n}$ do not converge because the terms that are being added don't go to zero fast enough, i.e. the change from $a_{n}$ to $a_{n+1}$ is too small.
A: "Is this principle the difference between a series converging or diverging?"
I don't think so.  For example, just consider a geometric series
$$\sum_{k=0}^\infty r^k$$
with $0<r<1$.  This will always converge.  However,
$$\sum_{k=n+1}^\infty r^k=\frac{r^{n+1}}{1-r}\ ,$$
and now


*

*if $r<\frac12$ then this is less than $r^k$: that is, any sum of subsequent terms is less than the $k$th term; whereas on the other hand

*if $r>\frac12$ then this is greater than $r^k$: that is, there exists a sum of (finitely many) subsequent terms which is greater than the $k$th term.


So as far as I can see, the observation has, in general, nothing to do with convergence.
A: Quick Reason: One is geometric and the other is a multiple of the harmonic series. 
Using the Integral Test,
$$\sum_{k=1}^{\infty} \frac{1}{2k} \ \ \textrm{divg/convg} \iff \lim_{k \to \infty} \int_{1}^{k} \frac{1}{2x} \ \textrm{dx}\ \ \textrm{divg/convg}$$
