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Proof:

let $s_n$ denote the $n^{th}$ partial sum of the series $\sum{a_n}$ and $t_n$ denote the $n^{th}$ partial sum of the series $\sum{|a_n|}$.

Since, the negative terms are finite

$\implies a_n \geq 0 $ $\forall n>K$

$\implies a_n = |a_n|$ $\forall n>K$

$\implies \forall m>n>K$: $s_m - s_n = t_m - t_n$

Now, since $\sum{a_n}$ is convergent, we can apply the cauchy criterion for series:

Let $\epsilon>0$ and $K \in \mathbb{N}$,

$|s_m - s_n| = |t_m - t_n| = |a_{n+1}+.....+a_m|<\epsilon$ $\forall m>n>K$

Can anyone verify this proof? I'm also unsure if I have used the cauchy criterion properly as it says: $\forall \epsilon>0$ , $ \exists K \in \mathbb{N}$ which I replaced with ''$\forall \epsilon>0$ and $K \in \mathbb{N}$" because I wanted to use the idea that the difference between partial sums of both series is equal after this K.

Thank you.

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That looks fine. Using the Cauchy criterion in this case is a bit overkill. List the negative elements as $a_{n_1}, \ldots, a_{n_K}$ and let $M = \sum_{k=1}^K \lvert a_{n_i} \rvert$. Then for $n \geq n_K$ we have $$ t_n = s_n + 2M \leq \sum_{n=0}^{\infty} a_n + 2M < \infty$$ So as a monotonically increasing sequence bounded above, it must converge.

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  • $\begingroup$ Sorry, I don't really understand your argument above. Also, are you sure it looks fine? Because we have to find a natural number opposite a given epsilon. And although it intuitively makes sense, I feel uneasy writing 'and for $K \in \mathbb{N}$'. $\endgroup$ – A.Asad Mar 31 '18 at 5:25
  • $\begingroup$ I got the part where you equated $t_n = s_n +2M$ and that, of course would be bounded above by the sum of the series + 2M. Then since $t_n$ is bounded, and it is an increasing sequence because all its terms are absolute values. Then, by MCT, it converges, which implies that the series converges absolutely. Is this the line of reasoning? $\endgroup$ – A.Asad Mar 31 '18 at 5:35
  • $\begingroup$ Yes, since the $t_n$ form an increasing sequence you just need to bound the sequence above. This question basically boils down to the result that if $\sum_{n = k}^{\infty} a_k$ converges then $\sum_{n=1}^{\infty} a_k$ converges which would be a slightly different way to prove it. $\endgroup$ – bitesizebo Mar 31 '18 at 8:22

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