If we have a series $\sum b_n$ which is bounded (its partial sums are bounded) and the terms go to zero, does it follow that it converges? I'm struggling to find a counterexample but don't know how to go forth with a proof.
edit: What other criteria is needed in our assumptions for this theorem to be true?
Consider $1-1/2-1/2 + 1/3+1/3+1/3 -1/4-1/4-1/4-1/4 +\cdots$ Its partial sums vary from $0$ to $1$ infinitely many times, hence are bounded, and the terms $\to 0.$ But the series diverges, precisely because of the oscillation of the partial sums.
No, it doesn't. Consider the bounded sequence $$1, \frac12,0, \frac13,\frac23,1,\frac 34,\frac24,\frac14,\dots$$ as the partial sums. Then the terms $b_n$, being $\pm\frac1k$, tend to $0$ as $n\to \infty$ but the series does not converge.
It need not converge. For example, let's build a sum $$ 1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\dots $$ by negating some of the terms of the harmonic series. We'll start by assigning a positive sign until the partial sum is at least $1$; then we'll assign negative signs until it is negative; then we'll assign positive signs until it's at least $1$ again; and so on.
The sequence of partial sums is bounded above by $2$ and below by $-1$. But the divergence of the harmonic series means we'll have infinitely many changes of sign, so it certainly cannot be Cauchy.
