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 $\sum\limits_{k=1}^\infty a_k=\prod\limits_{k=1}^\infty (1+a_k)\ne0$.
Best Answer
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) $$ \prod_{k=1}^\infty(1+a_k) = \frac{2}{c\pi} \sin\left(\frac{\pi(1+2c+\sqrt{1+4c^2})}{4}\right) \sin\left(\frac{\pi(1+2c-\sqrt{1+4c^2})}{4}\right) \\ \sum_{k=1}^\infty a_k = -c $$
There is a value $c$ where these are equal ... numerically $$ c \approx -0.73194187299064067002 $$