Wish to know the convergence of the double series {anbn}, where {an} and {bn} are both convergence series. The point is the product shown by anbn needs not to be termwise. There may be other ways one can take product, say a{i}b{n-i} where n runs from 1 to infinity and i also so. Some times it may so happen that with this type of product the double series may fail to converge but not by the term wise sense or vice-verse. I know the absolute convergence and/or conditional convergence may perhaps come into play for these cases? But how? I wish to know views from the experts or anybody interested on it.
Of course if an^2 and bn^2 converge then Cauchy Schwarz inequality tells us that anbn converges absolutely...
I should have posed the question differently, to make my point clear. Let sigma{an} and sigma{bn} be two given series, we can always form the double series sigmaf(m, n), where f(m, n) = ambn. For every arrangement g of f into a sequence G, we are led to a further series sigmaG(n) The convergence and sum of the series sigmaG(n) will depend on the arrangement of g. Different choices of g may yield different values of the product. But there is one very important case in which the terms f(m, n) are arranged diagonally to produce the Cauchy product and as @ Patrick_Sole has said the absolute convergence of series sigma(an) and sigma(bn) [ not the squared one, I think] implies convergence of the Cauchy product [ In fact Cauchy product arises quite naturally when we multiply two power series]. My point was about the other possible arrangement and their relation with convergence. I have simply no idea.... I would love no know on this.
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