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Let $\{a_n\}$ be a sequence of real numbers

The series $\sum_{k=1}^{\infty} a_k$ is summable if and only if for each $\epsilon > 0$ there is an index $N$ for which $|\sum_{k=n}^{n+m}a_k| < \epsilon$ for $n \geq N$ and any natural number $m$.

I am having a difficult time with this problem. Also does summable means converges.

Proof $\rightarrow$

I was thinking: if the series is summable then by the definition of Cauchy provided that for each $\epsilon > 0$ there is an index $N$ for which if $m,n \geq N$ then $| a_m - a_n| < \epsilon$ so it follows that $|\sum_{k=n}^{n+m}a_k| < \epsilon$ for $n \geq N$ and any natural number.

but i do not know if that is a correct approach

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    $\begingroup$ You know a sequence converges iff it is Cauchy. Apply this to the sequence of partial sums of your series. $\endgroup$ – zhw. Sep 26 '15 at 22:17
  • $\begingroup$ At the tail end of the statement, mention something like "and any natural number $m$", One is assuming each index is a natural number, but (to me) it would be better to emphasize the inequality is to hold for all positive integers $m$. $\endgroup$ – coffeemath Sep 26 '15 at 22:18
  • $\begingroup$ "$\sum_{n=0}^{\infty}a_n$ is summable " means that the sequence $(\sum_{n=0}^{n=m}a_n)_{m \in N}$ converges as $m \to \infty$. It is common, and quite acceptable, to say "$\sum_{n=0}^{\infty}a_n$ converges" to mean the same thing. $\endgroup$ – DanielWainfleet Sep 27 '15 at 1:41
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Define $x_n=\sum_{k=1}^na_k, n\in\mathbb{N}$ From the definition $\sum_{k=1}^{\infty}a_k$ is summable if and only if $(x_n)$ converges. We know that $(x_n)$ converges if and only if it is a Cauchy sequence. So $\sum_{k=1}^{\infty}a_k$ is summable if and only if $$\forall_{\varepsilon>0}\exists_{n_0\in\mathbb{N}}\forall_{a,b>n_0}\left|x_a-x_b\right|<\varepsilon$$ $$\left|\sum_{k=1}^{a}a_k-\sum_{k=1}^{b}a_k\right|<\varepsilon$$ Take $a=n+m, b=n-1$ which gives $$\left|\sum_{k=n}^{n+m}a_k\right|<\varepsilon$$

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    $\begingroup$ Note that you need $b=n-1$ because $\sum_{k=1}^a a_k-\sum_{k=1}^ba_k=\sum_{k=b+1}^a a_k$ for $a>b$. $\endgroup$ – Stephan Kulla Sep 26 '15 at 23:41

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