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If each term $a_k \geq 0$, then the series $\sum_{k=i}^{\infty}a_k$ is summable if and only if the sequence of a partial sums is bounded.

Proof: $\rightarrow$ Let $S_n$ be a partial sums

$$S_n = \sum_{k=i}^{\infty}a_k$$. such series is is of the form of a monotone sequence since $a_k \geq 0$ thus $S_n$ is bounded since every monotone sequence converges. and we kn0w that converges implies bounded.

I do not know for sure if this is reasonable. and also i do not know how to go to the other direction any idea that would be appreciated. thanks

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Assume $S_n < M, \forall n \geq 1$, then $S_n$ is an increasing sequence and is bounded above by $M$, hence converges to $L$. Thus $L = \displaystyle \lim_{n\to \infty} S_n = \displaystyle \sum_{i=1}^{\infty} a_i$. Thus the series is summable. Conversely, if the series is summable, and call the sum is $L$, then $L \geq S_n, \forall n \geq 1$, thus $\{S_n\}$ is bounded above by $L$.

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  • $\begingroup$ i think you need to edit it i can read it properly $\endgroup$ – user146269 Sep 27 '15 at 0:20
  • $\begingroup$ What part didn't you understand? $\endgroup$ – DeepSea Sep 27 '15 at 0:56
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If the series is bounded then even if you leave the first few terms and then add the series to the nttlh term the series will be convergent as the sequence of partial ss is convergent.

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Left to right: If the series sums to $s$, then no partial sum can be larger than $s$ since the $a_k$ are all positive. So $s$ is an upper bound for all partial sums.

Right to left: Again since all $a_k\geq 0$ note that the sequence of partial sums is monotone. Since it is assumed to be bounded, it must converge (by completeness: increasing sequences bounded above converge). So the series is summable.

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