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Homemanipulating convergent series

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# manipulating convergent series

The terms of the series in the following theorems are supposed to be either real or complex numbers.

###### Theorem 1.

Proof. The $n^{\mathrm{th}}$ partial sum of (1) has the limit

$\lim_{{n\to\infty}}\sum_{{j=1}}^{n}(a_{j}+b_{j})=\lim_{{n\to\infty}}\sum_{{j=1% }}^{n}a_{j}+\lim_{{n\to\infty}}\sum_{{j=1}}^{n}b_{j}=a\!+\!b.$ |

###### Theorem 2.

If the series $a_{1}+a_{2}+\cdots$ converges having the sum $a$ and if $c$ is any constant, then also the series

$\displaystyle ca_{1}+ca_{2}+\cdots$ | (2) |

converges and has the sum $ca$.

Proof. The $n^{\mathrm{th}}$ partial sum of (2) has the limit

$\lim_{{n\to\infty}}\sum_{{j=1}}^{n}ca_{j}=c\lim_{{n\to\infty}}\sum_{{j=1}}^{n}% a_{j}=ca.$ |

###### Theorem 3.

If the terms of any converging series

$\displaystyle a_{1}+a_{2}+a_{3}+\cdots$ | (3) |

are grouped arbitrarily without changing their order, then the resulting series

$\displaystyle(a_{1}+\cdots+a_{{m_{1}}})+(a_{{m_{1}+1}}+\cdots+a_{{m_{2}}})+(a_% {{m_{2}+1}}+\cdots+a_{{m_{3}}})+\cdots$ | (4) |

also converges and its sum equals to the sum of (3).

Proof. Since all the partial sums of (4) are simultaneously partial sums of (3), they have as limit the sum of the series (3).

Related:

SumOfSeries, MultiplicationOfSeries

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

40A05*no label found*26A06

*no label found*

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