multiplication of series

Theorem (Franz Mertens). If the series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ and ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ with real or complex converge and have the sums (http://planetmath.org/SumOfSeries) $A$ and $B$, respectively, and at least one of them converges absolutely, then also the series

$a_1{b}_1+(a_1{b}_2+a_2{b}_1)+(a_1{b}_3+a_2{b}_2+a_3{b}_1)+\mathrm{\cdots }$ (1)

is convergent and its is equal to $AB$.

Proof.  Denote the partial sums of the series  $A_n:=a_1+a_2+\mathrm{\cdots }+a_n$,  $B_n:=b_1+b_2+\mathrm{\cdots }+b_n$  and  $s_n:=a_1{b}_1+(a_1{b}_2+a_2{b}_1)+(a_1{b}_3+a_2{b}_2+a_3{b}_1)+\mathrm{\cdots }+(a_1{b}_n+a_2{b}_{n-1}+\mathrm{\cdots }+a_n{b}_1)$  for each $n$.  Then we have  ${lim}_{n\to \mathrm{\infty }}{A}_n=A$  and  ${lim}_{n\to \mathrm{\infty }}{B}_n=B$.  Suppose that e.g. the series $\sum a_n$ converges absolutely and that at least one $a_n$ is distinct from zero; so the   ${\sum }_{n=1}^{\mathrm{\infty }}|a_n|$ is a real positive number $M$.  Let $\epsilon$ be an arbitrary positive number.

Now we can write the identities

$AB=(A-A_n)B+a_{1}B+a_{2}B+\mathrm{\cdots }+a_{n}B$,

$s_n=a_1(b_1+b_2+\mathrm{\cdots }+b_n)+a_2(b_1+b_2+\mathrm{\cdots }+b_{n-1})+\mathrm{\cdots }+a_n{b}_1=a_1{B}_n+a_2{B}_{n-1}+\mathrm{\cdots }+a_n{B}_1$,

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

There is a positive number $n_1(\epsilon )$ such that    when  $n>n_1(\epsilon )$.  Then

(2)

The convergence of $\sum b_n$ implies that there is a number $n_2(\epsilon )$ such that    when  $n>n_2(\epsilon )$.  Thus we have

(3)

if  $n-k+1>n_2(\epsilon )$.  Because  ${lim}_{n\to \mathrm{\infty }}{B}_n=B$,  the numbers $|B_n|$ are bounded, i.e. there is a positive number $K$ such that for each $j$ we have    and consequently  $|B|\le K$.  It follows that    for every $j$.  We apply Cauchy criterion for convergence to the series ${\sum }_{n=1}^{\mathrm{\infty }}|a_n|$ getting a number $n_3(\epsilon )$ such that for each $m$, one has the inequality    if  $k>n_3(\epsilon )$.  Accordingly we obtain the estimation

(4)

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

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

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 by Cauchy where both series are

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

They converge by virtue of Leibniz test, but not absolutely (see the $p$-test (http://planetmath.org/PTest)).  In their product series

$$1-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})+(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})-+\mathrm{\cdots }$$

the absolute value of the $n^{\mathrm{th}}$ is  $1\cdot \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{n-1}}+\mathrm{\cdots }+\frac{1}{\sqrt{n}}\cdot 1$,  having $n$ 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  $y\le \frac{n+1}{2}$  and thus  $\frac{1}{y}\ge \frac{2}{n+1}$).  Because  ${lim}_{n\to \mathrm{\infty }}n\cdot \frac{2}{n+1}=2\ne 0$, the product series does not satisfy the necessary condition of convergence (http://planetmath.org/ThenA_kto0IfSum_k1inftyA_kConverges) and therefore the series diverges.

Title	multiplication of series
Canonical name	MultiplicationOfSeries
Date of creation	2014-10-31 20:23:29
Last modified on	2014-10-31 20:23:29
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	21
Author	pahio (2872)
Entry type	Theorem
Classification	msc 40A05
Synonym	Cauchy multiplication rule
Related topic	ManipulatingConvergentSeries
Related topic	AlternatingHarmonicSeries
Related topic	ErnstLindelof