| Version 7 |
Version 6 |
| A series $\sum_{i=0}^\infty a_i$ is \PMlinkid{convergent}{2311} iff for every $\varepsilon>0$ there is a number $N\in\mathbb{N}$ such that |
A series $\sum_{i=0}^\infty a_i$ is \PMlinkid{convergent}{2311} iff for every $\varepsilon>0$ there is a number $N\in\mathbb{N}$ such that |
| $$|a_{n+1}+a_{n+2}+\ldots+a_{n+p}|<\varepsilon$$ |
$$|a_{n+1}+a_{n+2}+\ldots+a_{n+p}|<\varepsilon$$ |
| holds for all $n>N$ and $p\geq1$. |
holds for all $n>N$ and $p\geq1$. |
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| \subsection*{Proof:} |
\subsection*{Proof:} |
| First define |
First define |
| $$s_n:=\sum_{i=0}^n a_i.$$ |
$$s_n:=\sum_{i=0}^n a_i.$$ |
| Now $(s_n)$ converges if and only if it is a Cauchy sequence, so if for every $\varepsilon>0$ there is a number $N$, such that for all $n,m>N$ holds: |
Now $(s_n)$ converges if and only if it is a Cauchy sequence, so if for every $\varepsilon>0$ there is a number $N$, such that for all $n,m>N$ holds: |
| $$|s_m-s_n|<\varepsilon.$$ |
$$|s_m-s_n|<\varepsilon.$$ |
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We can assume $m>n$ and thus set $m=n+p$. The he series is \PMlinkescapetext{convergent} iff
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We can assume $m>n$ and thus set $m=n+p$. The the series is \PMlinkescapetext{convergent} iff
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| $$|s_{n+p}-s_n|=|a_{n+1}+a_{n+2}+\ldots+a_{n+p}|<\varepsilon.$$ |
$$|s_{n+p}-s_n|=|a_{n+1}+a_{n+2}+\ldots+a_{n+p}|<\varepsilon.$$ |
