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Normally when I see this form of the product of infinite series:

$$ \left(\sum_{n=0}^\infty a_n\right)\left(\sum_{m=0}^\infty b_m\right) = \sum_{n=0}^\infty\sum_{m=0}^\infty a_nb_m $$

I see the claim that one of the series must be absolutely convergent (e.g. How to Multiply Two Infinite Series Correctly?). However, I have a simple proof that it is true for any convergent series. If this identity is true, it seems like it would be a proposition listed in most real analysis books, but it doesn't appear in any I've checked, which makes me suspect my proof may be invalid.

Here it is:

Suppose $ \sum_{n=0}^\infty a_n $ is a series of reals converging to $ x $, and suppose $ \sum_{m=0}^\infty b_m $ is a series of reals converging to $ y $.

For all $n \in \Bbb{N} $, we find that $ \sum_{m=0}^\infty a_nb_m $ is a convergent series with sum $a_ny$. And of course, $\sum_{n=0}^\infty a_ny$ is a convergent series with sum $xy$. Combining these two facts,

$$ \sum_{n=0}^\infty \sum_{m = 0}^\infty a_nb_m = \sum_{n=0}^\infty a_ny = xy $$

Is this valid? If so, why doesn't this appear in lists of properties of infinite series right alongside $ \sum_{n=0}^\infty a_n + \sum_{n=0}^\infty b_n = \sum_{n=0}^\infty (a_n + b_n)?$

Andrew Verras
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  • One thing: how do you define the product of two series? Why cannot we let it be $a_1b_1 + a_2b_1+ a_1b_2 + a_1b_3+ a_2b_2 + a_3 b_1 + \cdots$? Clearly there is more than 1 way to arrange the products $a_j b_k$. Simply speaking, why the product of two series is the double sums on the RHS? – xbh Nov 14 '18 at 03:58
  • I suppose I'm not defining the product of two infinite series at all. I'm simply demonstrating that the double series $\sum_{n=0}^\infty \sum_{m=0}^\infty a_nb_m $ converges and is equal to the product of the sums $x$ and $y$. – Andrew Verras Nov 14 '18 at 04:22
  • Then your demonstration is acceptable, but how does this be related to your title? – xbh Nov 14 '18 at 04:25
  • (As you noted) your equality is trivial. The Cauchy product is about $ \sum_{i=0}^{\infty} \sum_{j=0}^{i} a_{j} b_{i-j}$ – reuns Nov 14 '18 at 04:28
  • Until I thought through this proof, I thought the Cauchy product was the only way to multiply series. I'm just surprised I've never seen the equality listed as a property of convergent series. It seems less obvious than the formula for the sum of two series. – Andrew Verras Nov 14 '18 at 04:38
  • i would claim that this is the product of two series, not because I defined it to be, but simply because a convergent series represents a real number (its sum) and this is a series representation of the product of two (real) sums. So I suppose the expression I've discussed here is the "product" of two series (as follows from the definition of "product" of real numbers), and the "Cauchy product" is a new operation defined on series which ends up being equal to the "product" in the case one of the series is absolutely convergent. – Andrew Verras Nov 14 '18 at 05:06

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