2 Small corrections and clarifications

Thanks, Anthony, for finding this solution. I was completely at a loss for how to handle all the indices. If you don't mind, I would like to write down one version of the argument that you've given in full detail.

Claim: Under the hypotheses of the question $1 = k! \sum_{i_1 < \ldots < i_k} a_{i_1} \cdots a_{i_k} + O(\max |a_i|)$ where the error is non-negative.

The claim is true without an error when $k = 1$, and follows from induction. If we write $1 = (\sum a_i)^{k+1} = (\sum a_i) ( \sum a_i )^k$ The induction hypothesis allows us to write this product as $(\sum a_i)\cdot(k! \sum_{i_1 < \ldots < i_k} a_{i_1} + O(\max |a_i|) ) = (\sum a_i)\cdot (k! \sum_{i_1 < \ldots < i_k} a_{i_1} ) + O(\max |a_i| )$

If we now distribute out the product, we get the term we want $(k+1)! \sum_{i_1 < \ldots < i_k < i_{k+1} } a_{i_1} \cdots a_{i_{k+1} }$ from the products with no repeats and then an error coming from products with exactly one term repeated. Take whichever term is repeated and bound one copy of it in absolute value by $\max |a_i|$. Then the error is bounded by $\max |a_i| ( \sum |a_i| )^k = O(\max |a_i|)$.

Having this claim established and looking slightly more carefully at the dependence of the error on $k$ (the constant in the big O only grows like $C^k$), we also have prove the convergence that I was looking for (and we don't need non-negativity of the terms; just that $\sum |a_i|$ is bounded). In the non-negative case we can just observe the error is non-negative, so that the dominated convergence theorem applies (with respect to the finite measure $\frac{r^k}{k!}$), \frac{|r|^k}{k!}$), giving a small shortcut and a soft way to see the convergence without a rate. All credit goes to Anthony Quas for the idea; I just thought the induction was a fairly clear way to get the details all down. 1 Thanks, Anthony, for finding this solution. I was completely at a loss for how to handle all the indices. If you don't mind, I would like to write down one version of the argument that you've given in full detail. Claim: Under the hypotheses of the question$1 = k! \sum_{i_1 < \ldots < i_k} a_{i_1} \cdots a_{i_k} + O(\max |a_i|) $where the error is non-negative. The claim is true without an error when$k = 1$, and follows from induction. If we write$1 = (\sum a_i)^{k+1} = (\sum a_i) ( \sum a_i )^k$The induction hypothesis allows us to write this product as$(\sum a_i)\cdot(k! \sum_{i_1 < \ldots < i_k} a_{i_1} + O(\max |a_i|) ) = (\sum a_i)\cdot (k! \sum_{i_1 < \ldots < i_k} a_{i_1} ) + O(\max |a_i| ) $If we now distribute out the product, we get the term we want$(k+1)! \sum_{i_1 < \ldots < i_k < i_{k+1} } a_{i_1} \cdots a_{i_{k+1} }$from the products with no repeats and then an error coming from products with exactly one term repeated. Take whichever term is repeated and bound it by$\max |a_i|$. Then the error is bounded by$\max |a_i| ( \sum |a_i| )^k = O(\max |a_i|)$. Having this claim established and looking slightly more carefully at the dependence of the error on$k$(the constant in the big O only grows like$C^k$), we also have prove the convergence that I was looking for (and we don't need non-negativity of the terms; just that$\sum |a_i|$is bounded). In the non-negative case we can just observe the error is non-negative, so that the dominated convergence theorem applies (with respect to the finite measure$\frac{r^k}{k!}\$), giving a small shortcut and a soft way to see the convergence without a rate.

All credit goes to Anthony Quas for the idea; I just thought the induction was a fairly clear way to get the details all down.