Why won't a series converge if the limit of the sequence is 0?
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.
Do you think the series
$$1+\frac12+\frac12 + \frac14+\frac14+\frac14+\frac14 + \frac18 + \cdots$$
converges? Note, there are $2$ terms equal to $\frac12$, $4$ terms equal to $\frac14$, $8$ terms equal to $\frac18$ and so on, with $2^i$ terms equal to $\frac{1}{2^i}$ for each $i\in\mathbb N$.
You will probably agree that this series diverges. In fact, if you give me a number $m\in \mathbb N$, I can calculate exactly how many terms of the series you have to add together for the sum to reach $m$.
For example, it takes $1$ term to reach $1$, it takes $1+2=3$ terms to reach $2$, and then $1+2+4=7$ terms to reach $3$. You can show, with a simple inductive argument, that you will reach $m$ after
$$1+2+4+\dots + 2^{m-1}$$
terms, which is actually equal to $2^m-1$ and is certainly a finite number.
It's good to understand the concept why this series diverges. The thing is that yes, the terms go to $0$, but they don't do so "fast enough". The problem is than once the terms hit $\frac14$, they stick at that number for $4$ steps, long enough for the sum to increase by $1$.
And imagine what happens way way way down the line. The sum is equal to $\frac{1}{1024}$ for a whole $1024$ terms, for example. Sure, it will eventually fall to an even lower number, but it will stay on that number for even longer, again long enough for the whole sum to increase by $1$.
Side fact: the series I wrote down at the start has the bonus property that each term in the sequence is larger than the corresponding term of the sequence
$$1+\frac12+\frac13+\frac14+\frac15+\cdots$$
which is also known as the harmonic series and is the most famous divergent series. So, you now see that if you sum $2^m$ terms of the harmonic series, your sum will be equal to at least $m$ (and more, in fact).
However note that in the field of $p$-adic numbers $\mathbf Q_p$, a series $\sum_na_n$ converges if and only if the sequence $(a_n)$ tends to $0$.