multiplication of series

Theorem (Franz Mertens). If the series k=1ak and k=1bk with real or complex converge and have the sums ( A and B, respectively, and at least one of them converges absolutely, then also the series

a1b1+(a1b2+a2b1)+(a1b3+a2b2+a3b1)+ (1)

is convergentMathworldPlanetmath and its is equal to AB.

Proof.  Denote the partial sums of the series  An:=a1+a2++an,  Bn:=b1+b2++bn  and  sn:=a1b1+(a1b2+a2b1)+(a1b3+a2b2+a3b1)++(a1bn+a2bn-1++anb1)  for each n.  Then we have  limnAn=A  and  limnBn=B.  Suppose that e.g. the series an converges absolutely and that at least one an is distinct from zero; so the   n=1|an| is a real positive number M.  Let ε be an arbitrary positive number.

Now we can write the identities




There is a positive number n1(ε) such that  |A-An|<ε3(|B|+1)  when  n>n1(ε).  Then

|(A-An)B|=|A-An||B|<ε3(|B|+1)(|B|+1)=ε3. (2)

The convergence of bn implies that there is a number n2(ε) such that  |B-Bn|<ε3M  when  n>n2(ε).  Thus we have

|[]||a1||B-Bn|++|ak||B-Bn-k+1|<(|a1|++|ak|)ε3MMε3M=ε3 (3)

if  n-k+1>n2(ε).  Because  limnBn=B,  the numbers |Bn| are boundedPlanetmathPlanetmath, i.e. there is a positive number K such that for each j we have  |Bj|<K  and consequently  |B|K.  It follows that  |B-Bj||B|+|Bj|<K+K=2K  for every j.  We apply Cauchy criterion for convergence to the series n=1|an| getting a number n3(ε) such that for each m, one has the inequality|ak+1|++|am|<ε6K  if  k>n3(ε).  Accordingly we obtain the estimation

|ak+1(B-Bn-k)++an(B-B1)||ak+1||B-Bn-k|++|an||B-B1|<2Kε6K=ε3 (4)

which is valid when  k>n3(ε).

If we choose  n>max{n1(ε},n2(ε)+n3(ε)}  and k such that  nk>n3(ε)+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


They converge by virtue of Leibniz test, but not absolutely (see the p-test (  In their product series


the absolute valueMathworldPlanetmathPlanetmathPlanetmathPlanetmath of the nth is  11n+121n-1++1n1,  having n summands which all are greater than 2n+1 (this is seen when one looks at the half circle  y=xn+1-x  or  (x-n+12)2+y2=(n+12)2,  which shows that  yn+12  and thus  1y2n+1).  Because  limnn2n+1=20, the product series does not satisfy the necessary condition of convergence ( and therefore the series diverges.


  • 1 Ernst Lindelöf: Johdatus funktioteoriaan.  Mercatorin Kirjapaino Osakeyhtiö. Helsinki (1936).
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