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Let $ a_n$ be a sequence, so that $ \sum_{i = 1}^\infty a_n$ converges and $ a_k \ge 0 \forall k \in \mathbb{N} $

I want to prove the following property of convergent series:

$ \forall \epsilon \gt 0 \exists n_0 \in \mathbb{N} $ so that $ \sum_{i = n_0}^\infty <\epsilon $

However I think my proof is not very elegant (because I have probably overlooked something) and I think/hope there is a better way to show this.

(sketch of the proof)

Proof by contradiction: Now $ \exists \epsilon > 0 : \forall n_0 \in \mathbb{N} $ either $ \exists n_m \in \mathbb{N} $ so that: $ \sum_{i = n_0}^{n_m} > \epsilon $ (Case 1)

or $ \sum_{i = n_0}^{\infty} = \epsilon $ (Case 2)

Case 1 can be brought to contradiction, because $ \sum_{i = 1}^\infty a_n$ converges, while the series in Case 1 tends to infinity.

In case 2, there is obviously a $a_m \ne 0 $ for some m. Setting $a_0 = a_{m+1} $ gets one to: $ \sum_{i = n_0}^{\infty} = \epsilon - a_m < \epsilon $, which brings one to the wanted contradiction.

In my opinion even this sketch is far too long for such a simple or fundamental property of such a convergent series. Writing everything down clearly would extend the proof even further, thus I hope someone can show me a better/more elegant way to proof this. If there are mistakes in it, I will correct them as soon as possible. (Thx in advance)

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I think proving this by contradiction is unnecessary, as it follows almost immediately by definition. Let $(s_n)$ be the sequence of partial sums. Then since $a_k\geq0$, $(s_n)$ is increasing. Hence if the limit is $s$ then $$\sum_{i=n}^\infty a_i=s-s_{n-1}=|s-s_{n-1}|$$ so your result follows by definition of a limit.

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  • $\begingroup$ Yes, it is unnecessary. I didn't notice that.. $\endgroup$ – Imago Feb 7 '15 at 15:31

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