Why won't a series converge if the limit of the sequence is 0? Just thinking about it in terms of logic, shouldn't the series of a sequence whose limit as $n$ approaches infinity is 0 converge?

I know that the $n$th term test for divergence says that if a series is convergent, then the limit of its sequence is 0 and I also know there are some sequences for which it has been "proven" that their series does not converge even though the sequence converges to 0, but I just don't believe these tests. If we stretch $n$ out to infinity and the terms are approaching 0, then how is it possible for the sum of the terms to "overflow" and diverge if the terms are becoming negligibly small?
A very easy counterexample would be
$$ 1, \underbrace{\frac12, \frac12}_{2\text{ halves}},
\underbrace{\frac13, \frac13, \frac13}_{3\text{ thirds}},
\underbrace{\frac14, \frac14, \frac14, \frac14}_{4\text{ fourths}},
\underbrace{\frac15, \frac15, \frac15, \frac15, \frac15}_{5\text{ fifths}},
\ldots $$
This sequence clearly converges to $0$, but if you try to sum it, it should be obvious that it has partial sums as large as you'd like them to be -- so the series diverges.

Try whichever argument you have in mind for believing that the series should converge, and attempt to figure out why it doesn't work for this one.
Just look at the problem from the other end.

Take a function like $f(x) = \sqrt{x}$ or $\log x$ - they grow infinitely with the $x$ growth, though their growth ratio decreases (their first derivatives tend to zero).

Consider the function increments for equal steps of argument, say, for arguments natural:
$$\Delta f = f(n+1) - f(n)\quad \text{ for } n\in\mathbb N^+$$
As the $f$ function grows slower and slower, these deltas get smaller and smaller for growing $n$, and they actually tend to zero: for arbitrarily small $\varepsilon &gt; 0$ there exists $n$ big enough, for which $\Delta f &lt; \varepsilon$.

Now $\Delta f$ defines a sequence convergent to zero, whose partial sums are values of $\sqrt n$, that is a function growing with no bound, hence a divergent series.