PlanetMath: manipulating convergent series

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manipulating convergent series

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

Theorem 1 If the series $a_1+a_2+\cdots$ and $b_1+b_2+\cdots$ converge and have the sums

and

, respectively, then also the series

$\displaystyle (a_1+b_1)+(a_2+b_2)+\cdots$

(1)

converges and has the sum $a\!+\!b$ .

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

$\displaystyle \lim_{n\to\infty}\sum_{j = 1}^n(a_j+b_j) = \lim_{n\to\infty}\sum_{j = 1}^na_j+\lim_{n\to\infty}\sum_{j = 1}^nb_j = a\!+\!b.$

Theorem 2 If the series $a_1+a_2+\cdots$ converges having the sum

and if

is any constant, then also the series

$\displaystyle ca_1+ca_2+\cdots$

(2)

converges and has the sum

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

$\displaystyle \lim_{n\to\infty}\sum_{j = 1}^nca_j = c\lim_{n\to\infty}\sum_{j = 1}^na_j = ca.$

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).

"manipulating convergent series" is owned by pahio.

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Cross-references: limit, partial sum, sums, converge, complex numbers, real, series
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This is version 9 of manipulating convergent series, born on 2004-11-23, modified 2006-10-14.
Object id is 6517, canonical name is ManipulatingConvergentSeries.
Accessed 3275 times total.

Classification:

AMS MSC:	40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)
	26A06 (Real functions :: Functions of one variable :: One-variable calculus)

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