# multiplication of series

Theorem (Franz Mertens). If the series $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=1}^{\infty}b_{k}$ 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

 $\displaystyle a_{1}b_{1}\!+\!(a_{1}b_{2}\!+\!a_{2}b_{1})\!+\!(a_{1}b_{3}\!+\!a% _{2}b_{2}\!+\!a_{3}b_{1})\!+\cdots$ (1)

is convergent  and its is equal to $AB$.

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  $s_{n}:=a_{1}b_{1}+(a_{1}b_{2}+a_{2}b_{1})+(a_{1}b_{3}+a_{2}b_{2}+a_{3}b_{1})+% \cdots+(a_{1}b_{n}+a_{2}b_{n-1}+\cdots+a_{n}b_{1})$  for each $n$.  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 $a_{n}$ is distinct from zero; so the   $\sum_{n=1}^{\infty}|a_{n}|$ is a real positive number $M$.  Let $\varepsilon$ be an arbitrary positive number.

Now we can write the identities

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

$s_{n}=a_{1}(b_{1}+b_{2}+\cdots+b_{n})+a_{2}(b_{1}+b_{2}+\cdots+b_{n-1})+\cdots% +a_{n}b_{1}=a_{1}B_{n}+a_{2}B_{n-1}+\cdots+a_{n}B_{1}$,

$AB\!-\!s_{n}=(A-A_{n})B+a_{1}(B-B_{n})+a_{2}(B-B_{n-1})+\cdots+a_{n}(B-B_{1})$
$=(A-A_{n})B+[a_{1}(B-B_{n})+a_{2}(B-B_{n-1})+\cdots+a_{k}(B-B_{n-k+1})]\\ +a_{k+1}(B-B_{n-k})+\cdots+a_{n}(B-B_{1})$.

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

 $\displaystyle|(A\!-\!A_{n})B|=|A\!-\!A_{n}|\cdot|B|<\frac{\varepsilon}{3(|B|\!% +\!1)}(|B|\!+\!1)=\frac{\varepsilon}{3}.$ (2)

The convergence of $\sum b_{n}$ implies that there is a number $n_{2}(\varepsilon)$ such that  $|B\!-\!B_{n}|<\frac{\varepsilon}{3M}$  when  $n>n_{2}(\varepsilon)$.  Thus we have

 $\displaystyle|[\ldots]|\leq|a_{1}|\!\cdot\!|B\!-\!B_{n}|\!+\cdots+\!|a_{k}|\!% \cdot\!|B\!-\!B_{n-k+1}|<(|a_{1}|\!+\cdots+\!|a_{k}|)\frac{\varepsilon}{3M}% \leq M\!\cdot\!\frac{\varepsilon}{3M}=\frac{\varepsilon}{3}$ (3)

if  $n\!-\!k\!+\!1>n_{2}(\varepsilon)$.  Because  $\lim_{n\to\infty}B_{n}=B$,  the numbers $|B_{n}|$ are bounded  , i.e. there is a positive number $K$ such that for each $j$ we have  $|B_{j}|  and consequently  $|B|\leq K$.  It follows that  $|B\!-\!B_{j}|\leq|B|\!+\!|B_{j}|  for every $j$.  We apply Cauchy criterion for convergence to the series $\sum_{n=1}^{\infty}|a_{n}|$ getting a number $n_{3}(\varepsilon)$ such that for each $m$, one has the inequality$|a_{k+1}|+\cdots+|a_{m}|<\frac{\varepsilon}{6K}$  if  $k>n_{3}(\varepsilon)$.  Accordingly we obtain the estimation

 $\displaystyle|a_{k+1}(B\!-\!B_{n-k})\!+\cdots+\!a_{n}(B\!-\!B_{1})|\leq|a_{k+1% }||B\!-\!B_{n-k}|\!+\cdots+\!|a_{n}||B\!-\!B_{1}|<2K\!\cdot\!\frac{\varepsilon% }{6K}=\frac{\varepsilon}{3}$ (4)

which is valid when  $k>n_{3}(\varepsilon)$.

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

 $|AB\!-\!s_{n}|<\frac{\varepsilon}{3}\!+\!\frac{\varepsilon}{3}\!+\!\frac{% \varepsilon}{3}=\varepsilon.$

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 by Cauchy where both series are

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

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

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

the absolute value     of the $n^{\mathrm{th}}$ is  $1\!\cdot\!\frac{1}{\sqrt{n}}\!+\!\frac{1}{\sqrt{2}}\frac{1}{\sqrt{n-1}}\!+% \cdots+\frac{1}{\sqrt{n}}\!\cdot 1\!$,  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  $y\leq\frac{n+1}{2}$  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 (http://planetmath.org/ThenA_kto0IfSum_k1inftyA_kConverges) and therefore the series diverges.

## References

• 1 Ernst Lindelöf: Johdatus funktioteoriaan.  Mercatorin Kirjapaino Osakeyhtiö. Helsinki (1936).
Title multiplication of series MultiplicationOfSeries 2014-10-31 20:23:29 2014-10-31 20:23:29 pahio (2872) pahio (2872) 21 pahio (2872) Theorem msc 40A05 Cauchy multiplication rule ManipulatingConvergentSeries AlternatingHarmonicSeries ErnstLindelof