The point is that just because a series is bounded, this doesn't mean it converges. 1 - 1 + 1 - 1 + ... is one such example. Richard's is another one. Here is the definition determining whether a series $a_1+a_2+\dots$converges:

Let $s_n=a_1+\dots +a_n$. Then the series $\sum _{i=1}^{\infty }{a}_i$ converges to $L$ if and only if for every $\epsilon >0$ there exists an integer $N$ such that

$-\epsilon <s_n-L<\epsilon$

for all $n>N$.

In other words the series gets to within an arbitrarily close proximity to $L$ and then stays there forever, provided you look at the partial sum sufficiently far across. If you have a look at Richard's counterxample, you can see that this condition is not satisfied.

If you have a convergent series, then if you split up the terms into several parts and shove them in then it won't necessarily converge. So while:

0 + 0 + 0 + ...

converges, and so does:

(1 - 1) + (1 - 1) + ...

(exactly the same) it is not OK to ungroup the brackets:

1 - 1 + 1 - 1 + ...

In the same way, if $a_i$ is a decreasing positive sequence then sure:

(a0 - a1) + (a2 - a3) + ...

and

a0 + (-a1 + a2) + (-a3 + a4) + ...

both individually converge (because one is increasing and bounded above and the other is decreasing and bounded below) but if you remove the brackets it won't necessarily converge. (If $a_i\to 0$ then by the alternating series test, which I'll prove if you like although it's not hard, it does converge.) If the $n$th partial sum of any series doesn't tend to 0 then the series doesn't converge. But the converse isn't always true.

Finally, I should point out that in general, a sequence being bounded and having its terms tending to 0 still isn't a sufficient condition for convergence. For instance if the terms $a_i$ in the series are:

1

-0.5

-0.5

0.25

0.25

0.25

0.25

-0.125

-0.125

-0.125

-0.125

-0.125

-0.125

-0.125

-0.125

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625

0.0625