PlanetMath: multiplication of series

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (193)
Orphanage
Unclass'd
Unproven (498)
Corrections (23)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

multiplication of series

(Theorem)

Theorem 1 If the series $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty b_k$ with real or complex terms converge and have the sums

and

, respectively, and at least one of them converges absolutely, then also the series

$\displaystyle a_1b_1\!+\!(a_1b_2\!+\!a_2b_1)\!+\!(a_1b_3\!+\!a_2b_2\!+\!a_3b_1)\!+\cdots$

(1)

is convergent and its sum is equal to

Proof. Denote the partial sums of the series $A_n := a_1+a_2+\cdots+a_n$ , $B_n := b_1+b_2+\cdots+b_n$ and $s_n := a_1b_1+(a_1b_2+a_2b_1)+(a_1b_3+a_2b_2+a_3b_1)+\cdots +(a_1b_n+a_2b_{n-1}+\cdots+a_nb_1)$ for each . Then we have $\lim_{n\to\infty}A_n = A$ and $\lim_{n\to\infty}B_n = B$ . Suppose that e.g. the series $\sum a_n$ converges absolutely and that at least one is distinct from zero; so the sum $\sum_{n=1}^\infty\vert a_n\vert$ is a real positive number . Let $\varepsilon$ be an arbitrary positive number.

Now we can write the identities

$AB = (A-A_n)B+a_1B+a_2B+\cdots+a_nB$ ,

$s_n = a_1(b_1+b_2+\cdots+b_n)+a_2(b_1+b_2+\cdots+b_{n-1})+\cdots+a_nb_1 = a_1B_n+a_2B_{n-1}+\cdots+a_nB_1$ ,

$AB\!-\!s_n = (A-A_n)B+a_1(B-B_n)+a_2(B-B_{n-1})+\cdots+a_n(B-B_1)$
$= (A-A_n)B+[a_1(B-B_n)+a_2(B-B_{n-1})+\cdots+a_k(B-B_{n-k+1})]\ +a_{k+1}(B-B_{n-k})+\cdots+a_n(B-B_1)$ .

There is a positive number $n_1(\varepsilon)$ such that $\vert A\!-\!A_n\vert < \frac{\varepsilon}{3(\vert B\vert+1)}$ when $n > n_1(\varepsilon)$ . Then

$\displaystyle \vert(A\!-\!A_n)B\vert = \vert A\!-\!A_n\vert\cdot\vert B\vert < ... ...varepsilon}{3(\vert B\vert\!+\!1)}(\vert B\vert\!+\!1) = \frac{\varepsilon}{3}.$

(2)

The convergence of $\sum b_n$ implies that there is a number $n_2(\varepsilon)$ such that $\vert B\!-\!B_n\vert < \frac{\varepsilon}{3M}$ when $n > n_2(\varepsilon)$ . Thus we have

$\displaystyle \vert[\ldots]\vert \leq \vert a_1\vert\!\cdot\!\vert B\!-\!B_n\ve... ...{\varepsilon}{3M} \leq M\!\cdot\!\frac{\varepsilon}{3M} = \frac{\varepsilon}{3}$

(3)

if $n\!-\!k\!+\!1 > n_2(\varepsilon)$ . Because $\lim_{n\to\infty}B_n = B$ , the numbers $\vert B_n\vert$ are bounded, i.e. there is a positive number

such that for each

we have $\vert B_j\vert < K$ and consequently $\vert B\vert \leq K$ . It follows that $\vert B\!-\!B_j\vert \leq \vert B\vert\!+\!\vert B_j\vert < K\!+\!K = 2K$ for every

. We apply Cauchy criterion for convergence to the series $\sum_{n=1}^\infty\vert a_n\vert$ getting a number $n_3(\varepsilon)$ such that for each

, one has the inequality $\vert a_{k+1}\vert+\cdots+\vert a_m\vert < \frac{\varepsilon}{6K}$ if $k > n_3(\varepsilon)$ . Accordingly we obtain the estimation

$\displaystyle \vert a_{k+1}(B\!-\!B_{n-k})\!+\cdots+\!a_n(B\!-\!B_1)\vert \leq ... ...vert B\!-\!B_1\vert < 2K\!\cdot\!\frac{\varepsilon}{6K} = \frac{\varepsilon}{3}$

(4)

which is valid when $k > n_3(\varepsilon)$ .

If we choose $n > \max\{n_1(\varepsilon\},\, n_2(\varepsilon)\!+\!n_3(\varepsilon)\}$ and such that $n \geq k > n_3(\varepsilon)\!+\!1$ , then the inequalities (2), (3) and (4) are satisfied, ensuring that

$\displaystyle \vert AB\!-\!s_n\vert < \frac{\varepsilon}{3}\!+\!\frac{\varepsilon}{3}\!+\!\frac{\varepsilon}{3} = \varepsilon.$

This means that the assertion of the theorem has been proved.

Remark. The mere convergence of both series does not suffice for convergence of (1). This is seen in the following example where both series are

$\displaystyle 1\!-\!\frac{1}{\sqrt{2}}\!+\!\frac{1}{\sqrt{3}}\!-+\cdots$

They converge by virtue of Leibniz test, but not absolutely (see the

-test). In their product series

$\displaystyle 1\!-\!(\frac{1}{\sqrt{2}}\!+\!\frac{1}{\sqrt{2}})\!+\!(\frac{1}{\... ...{3}}\!+\!\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\!+\!\frac{1}{\sqrt{3}})\!-+\cdots$

the absolute value of the $n^\mathrm{th}$ term is $1\!\cdot\!\frac{1}{\sqrt{n}}\!+\!\frac{1}{\sqrt{2}}\frac{1}{\sqrt{n-1}}\!+\cdots +\frac{1}{\sqrt{n}}\!\cdot1\!$ , having

summands which all are greater than $\frac{2}{n+1}$ (this is seen when one looks at the half circle $y = \sqrt{x}\sqrt{n\!+\!1\!-\!x}$ or $(x\!-\!\frac{n+1}{2})^2\!+\!y^2 = (\frac{n+1}{2})^2$ , which shows that $y \leq \frac{n+1}{2}$ and thus $\frac{1}{y} \geq \frac{2}{n+1}$ ). Because $\lim_{n\to\infty}n\cdot\frac{2}{n+1} = 2 \neq 0$ , the product series does not satisfy the necessary condition of convergence and therefore the series diverges.

Bibliography

1: ERNST LINDELÖF: Johdatus funktioteoriaan. Mercatorin Kirjapaino Osakeyhtiö. Helsinki (1936).

"multiplication of series" is owned by pahio.

(view preamble)

Other names:

Cauchy multiplication rule

This object's parent.

Attachments:: example of Cauchy multiplication rule (Example) by pahio
Abel's multiplication rule for series (Theorem) by pahio

Log in to rate this entry.
(view current ratings)

Cross-references: diverges, circle, absolute value, product, Leibniz test, inequality, Cauchy criterion for convergence, bounded, implies, identities, number, positive, partial sums, proof, convergent, converges absolutely, converge, complex, real, series
There are 6 references to this entry.

This is version 16 of multiplication of series, born on 2005-10-10, modified 2008-03-26.
Object id is 7431, canonical name is MultiplicationOfSeries.
Accessed 5931 times total.

Classification:

AMS MSC:

40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)

Pending Errata and Addenda

None.[ View all 1 ]

Discussion

forum policy

No messages.

Interact