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Suppose $a_n$ is a sequence in $\mathbb R$ and $S_n=\sum_{k=1}^na_k$ when $S_n$ has bound.

Then does $\sum_{k=1}^\infty a_k$ converge?

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No: the series $\sum_n(-1)^n$ is a counter-example.

However, if the series has non-negative terms, and if the partial sums are bounded, the series converge by the monotone convergence theorem.

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  • $\begingroup$ Because the partial sums will form an increasing sequence and hence, if bounded, will converge to the supremum? $\endgroup$ – A.Asad Mar 30 '18 at 12:05
  • $\begingroup$ @A.Asad: Yes, exactly. $\endgroup$ – Bernard Mar 30 '18 at 12:09
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Take $a_n=(-1)^n$. As you can see, $S_n$ is not convergent.

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No.

The series converges if and only if $S_n$ has a limit as $n\to\infty$. Having a limit does not follow from being bounded. For any sequence $(S_n)$ you can find the corresponding sequence $(a_n)$ by letting $a_n=S_n-S_{n-1}$.

Now you can take your favorite bounded sequence $(S_n)$ with no limit and define $a_n$ like above. Then your conditions are satisfied but the series won't converge.

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Well, the series is not convergent necessarily but since $S_n$ is bounded, there is a convergent subsequence $S_{n_k}$ of $S_n$. In other words, a "sub-series" of $\sum_{n}a_n $ converges.

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