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
Best Answer
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 $$\left|\sum_{k=n}^{n+m}a_k\right|<\varepsilon$$