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I want to show \begin{equation*} \sum_{n=1}^\infty f(n)\sum_{m=1}^\infty g(m)=\sum_{n=1}^\infty \sum_{m=1}^\infty f(n)g(m)\end{equation*} Do I have to invoke the conditions on Fubini's theorem? Or is this always true?

I tried this proof, which does not require absolute convergence, just convergence. \begin{align*} \sum_{n=1}^\infty f(n)\sum_{m=1}^\infty g(m)&=\lim_{N\rightarrow\infty}\sum_{n=1}^Nf(n)\sum_{m=1}^\infty g(m)\\ &=\lim_{N\rightarrow\infty}\left[f(1)\sum_{m=1}^\infty g(m)+...+f(N)\sum_{m=1}^\infty g(m)\right]\\ &=\lim_{N\rightarrow\infty}\left[f(1)\lim_{M\rightarrow\infty}\sum_{m=1}^M g(m)+...+f(N)\lim_{M\rightarrow\infty}\sum_{m=1}^M g(m)\right]\\ &=\lim_{N\rightarrow\infty}\left[\lim_{M\rightarrow\infty}\sum_{m=1}^M f(1)g(m)+...+\lim_{M\rightarrow\infty}\sum_{m=1}^M f(N)g(m)\right]\\ &=\lim_{N\rightarrow\infty}\lim_{M\rightarrow\infty}\left[\sum_{m=1}^M f(1)g(m)+...+\sum_{m=1}^M f(N)g(m)\right]\\ &=\lim_{N\rightarrow\infty}\lim_{M\rightarrow\infty}\sum_{m=1}^N\sum_{m=1}^M f(n)g(m)\\ &=\lim_{N\rightarrow\infty}\sum_{m=1}^N\lim_{M\rightarrow\infty}\sum_{m=1}^M f(n)g(m)\\ &=\sum_{n=1}^\infty\sum_{m=1}^\infty f(n) g(m) \end{align*}

KonKan
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telemaco
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  • I believe your proof is ok as long as all three summations converge. – KonKan Oct 12 '16 at 00:07
  • I'm curious if this still works if the sums diverge to infinity – telemaco Oct 12 '16 at 00:18
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    Generally no, in the following sense: there may be a situation in which the two sums in the LHS are convergent (but not absolutely convergent; imagine for example alternating signs in each of $f(n)$, $g(n)$) and the sum in the RHS has all its terms positive. In this case it will be divergent, since the summations on $|f(n)|$, $|g(m)|$ will each be divergent. – KonKan Oct 12 '16 at 00:25

1 Answers1

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If what you mean is $$ \begin{equation*} \bigg(\sum_{n=1}^\infty f(n)\bigg)\cdot\bigg(\sum_{m=1}^\infty g(m)\bigg)=\sum_{n=1}^\infty \sum_{m=1}^\infty f(n)g(m) \end{equation*} $$ this is always valid, as long as the summations $\sum_{n=1}^\infty f(n)$, $ \ \sum_{m=1}^\infty g(m)$ are absolutely convergent or more generally as long as all three summations are convergent: it is a kind of generalized statement of the distributivity of multiplication over addition.

KonKan
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  • Yes, I mean multiplication, that is for example, in the case where $\sum f(n)=2$ and $\sum g(m)=3$, then $\sum f(n)\sum g(m)=6$. How can I prove that? Just using limits should work? – telemaco Oct 11 '16 at 23:19
  • I believe you need all the sums to be absolutely convergent. Distributivity is written in terms of finite sums and products. I don't have a handy counterexample with conditionally convergent sums. – Ross Millikan Oct 11 '16 at 23:32
  • Hmm I think you may have a point here. I mean if the series are absolutely convergent then it seems clear, otherwise it needs a little more care. Thanks for noticing that. – KonKan Oct 11 '16 at 23:39
  • However, i think it is not necessary to demand absolute convergence for all three sums. – KonKan Oct 11 '16 at 23:44
  • @KonKan Please see my attempted proof in the edit. I don't use absolute convergence, just convergence, – telemaco Oct 11 '16 at 23:49
  • Yes I was studying it right now. Looks ok to me. So, you simply assume convergence for alll three summations. This is what I had in mind when I said "a kind of generalized statement of the distributivity". – KonKan Oct 11 '16 at 23:57