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I have been given the following definition: $\rule{17cm}{0.4pt}$

Let $\{a_n\}$ be a sequence in $\mathbb{R}$. The series: $$\sum_{n=0}^\infty a_n$$ is $\textbf{convergent}$ if the sequence $\{s_m\}$ of the $\textbf{partial sums}$ $$s_m=\sum_{n=0}^m a_n$$ converges, that is for all $\varepsilon >0$ there exists $N=N(\varepsilon)\in \mathbb{N},$ such that $$\left| s_m-s_k\right| = \left| \sum_{n=k+1}^m a_n \right| <\varepsilon$$ for all $m>k\geq N$.

$\{a_n\}$ $\textbf{converges absolutely}$ if the series: $$\sum_{n=0}^\infty |a_n|$$ converges. $\rule{17cm}{0.4pt}$

I was a bit confused about what it means by partial sums, and why, if the series of partial sums converges, that we know the $\{a_n\}$ converges.

I am thinking that for any finite m, $s_m$ would be a sub-sequence of $a_n$, and i think I am right in saying that if a subsequence is convergent then the sequence must also be convergent. This is a complete guess though, If someone could let me know if I am on the right track in understanding this, that'd be great. Thanks.

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  • $\begingroup$ The point is that we can't really sum infinitely many numbers, so instead we sum every finite sequence and then take limit in other words $$\sum_{n=1}^\infty a_n = \lim_{N\rightarrow\infty} \sum_{n=1}^N a_n$$ The partial sums are exactly $s_N = \sum_{n=1}^N a_n$. (This is not a sub-sequence of $a_n$!!!) $\endgroup$ – Yanko Dec 28 '17 at 18:39
  • $\begingroup$ Note correct MathJax usage, as in my edits to this question. In particular I changed $\sum\limits_{n=0}^\inf$ to $\sum\limits_{n=0}^\infty$ put the $\{$curly braces$\}$ inside rather than outside of MathJax to get proper spacing an matching of fonts, thus: $$ \{a_n\} $$ $\endgroup$ – Michael Hardy Dec 28 '17 at 19:05
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You have a sequence $a_1,a_2,a_3,a_4,a_5,\ldots$

The sequence of partial sums is \begin{align} s_1 & = a_1 \\[6pt] s_2 & = a_1 + a_2 \\[6pt] s_3 & = a_1 + a_2 + a_3 \\[6pt] s_4 & = a_1 + a_2 + a_3 + a_4 \\ & \,\,\,\,\vdots \\ s_n & = a_1 + \quad\cdots\cdots\cdots\quad + a_n \\ & \,\,\,\,\vdots \end{align} The sequence $\{s_n\}$ converges (usually to a nonzero number) only if the sequence $\{a_n\}$ converges to $0$ (and in many cases does not converge then).

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In general, $(s_n)_{n\in\mathbb N}$ is not a sub-sequence of $(a_n)_{n\in\mathbb N}$. However, if $\lim_{n\to\infty}s_n=l$, then $\lim_{n\to\infty}s_n-s_{n-1}=0$, which means that $\lim_{n\to\infty}a_n=0$ (because $a_n=s_n-s_{n-1}$).

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