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Is there a sequence of unique terms $\{a_k\}$ such that $\sum\limits_{k=1}^\infty a_k=\prod\limits_{k=1}^\infty (1+a_k)\ne0$ ?

I specify "unique terms" to rule out trivial cases like $a_1=2$, $a_2=-0.5$ and the other terms all $0$.

I specify "$\ne0$" to rule out trivial cases like $a_1=-1$, $a_2=1$ and the other terms sum to $0$.

Certainly some of the terms must be negative, because the expansion of the product contains the series in addition to other terms.

Context:

From another question, we see that $\int_0^\infty\left(\frac{\sin x}{x}\right)^2\mathrm dx=\dfrac{\pi}{2}$, and it is conjectured that $\prod\limits_{k=1}^\infty\left(1+\int_{k}^{k+1}\left(\frac{\sin (\pi x)}{x}\right)^2\mathrm dx\right)=\dfrac{\pi}{2}$.

The functions inside each integral are not exactly the same. But anyway, that question made me wonder if there can be a sequence of unique terms $\{a_k\}$ such that $\sum\limits_{k=1}^\infty a_k=\prod\limits_{k=1}^\infty (1+a_k)\ne0$.

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    $\begingroup$ Ah, my mistake. Then it's easy to get a positive answer over the complex numbers, though I'm not sure of a real example. For example, letting $c$ be a solution to $\cos(c) = -c^2/2$ (e.g. $c \approx 1.49584 + 1.62221i$) and $a_k = \frac{-c^2}{\pi^2 (n-\tfrac 12)^2}$ we get $\sum_{k=1}^\infty a_k = \prod_{k=1}^\infty (1+a_k) = \cos(c) \approx 0.197015 - 2.42657 i$, coming from the infinite product expansion of cosine. The trick is basically to consider a family of infinite products and equate it to the associated family of series and use continuity to find a solution. $\endgroup$ Nov 11, 2023 at 16:24

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There are many many of these sequences. Just some fidding produced this example...

$$ a_k = \frac{(-1)^k(2k+1)c}{k(k+1)} = (-1)^k\left(\frac{c}{k}+\frac{c}{k+1}\right) $$ where $c$ is a certain constant. I computed (using Maple) $$ \prod_{k=1}^\infty(1+a_k) = \frac{2}{c\pi} \sin\left(\frac{\pi(1+2c+\sqrt{1+4c^2})}{4}\right) \sin\left(\frac{\pi(1+2c-\sqrt{1+4c^2})}{4}\right) \\ \sum_{k=1}^\infty a_k = -c $$ A
There is a value $c$ where these are equal ... numerically $$ c \approx -0.73194187299064067002 $$

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Self-answering. This is easier than I thought. Here is an example given in closed form.

Let $a_k=\dfrac{1}{k^2}$ for $k\ge2$. Now we will calculate $a_1$.

$\sum\limits_{k=1}^\infty a_k=a_1+\frac{\pi^2}{6}-1$ (proof)

$\prod\limits_{k=1}^\infty (1+a_k)=(1+a_1)\prod\limits_{k=2}^\infty \left(1+\frac{1}{k^2}\right)=(1+a_1)\frac{\sinh{\pi}}{2\pi}$ (proof)

Setting the series and product equal yields $a_1=\dfrac{\frac{\sinh{\pi}}{2\pi}-\frac{\pi^2}{6}+1}{1-\frac{\sinh{\pi}}{2\pi}}\approx −1.42368666869$.

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    $\begingroup$ Indeed, and that would work with any convergent series as long as $\prod_{k=2}^\infty (1+a_k) \ne 1$. $\endgroup$
    – Martin R
    Nov 11, 2023 at 18:03

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