Investigate whether sum from k=1 to infinity a_k=product

Investigate whether $\displaystyle\sum_{k=1}^{\infty}a_k=\prod_{k=1}^{\infty}(1-a_k)$ is possible for a sequence with $0

Explanation of Method

Write $S=\sum a_k$ and $P=\prod(1-a_k)$. Two classical, sharp bounds pin down $P$ in terms of $S$ alone: Weierstrass's product inequality $P\ge1-S$, and a matching asymptotic upper bound $P

Bound below using Weierstrass's inequality $\prod(1-a_k)\ge1-\sum a_k$.
Bound above using the "infinitely thin spreading" limit $\prod(1-a_k)
Combine both bounds under $S=P$ to get the sharp range $\tfrac12\le S<\Omega$, and verify by explicit construction.

Step 1 · The lower bound (Weierstrass's inequality)

For $a,b\in[0,1)$: $(1-a)(1-b)=1-a-b+ab\ge1-a-b$ since $ab\ge0$. By induction, for any finite (or, by taking limits, infinite) collection with $\sum a_k=S<\infty$: $$ \prod_{k=1}^{\infty}(1-a_k)\ \ge\ 1-\sum_{k=1}^{\infty}a_k = 1-S, $$ with equality exactly when at most one $a_k$ is nonzero. If $S=P$, this forces $S\ge1-S$, i.e. $\boxed{S\ge\tfrac12}$.

Step 2 · The upper bound

For fixed $S$, spreading the mass over $M\to\infty$ equal terms $a_k=S/M\to0$ gives $$ P=\left(1-\frac SM\right)^M\longrightarrow e^{-S}\qquad(M\to\infty), $$ and this is the supremum of $P$ over all distributions with $\sum a_k=S$ (concentrating mass, by Step 1, only ever decreases $P$ toward $1-S$). So $P

Step 3 · The sharp range, and constructions

Combining both bounds, equality $S=P$ can only occur for $$ \boxed{\tfrac12\ \le\ S\ <\ \Omega\approx0.567143.} $$ At $S=\tfrac12$: the single-term sequence $a_1=\tfrac12$ (all other $a_k=0$) trivially gives $S=P=\tfrac12$. For $S$ strictly inside $(\tfrac12,\Omega)$, a mixed construction works: e.g. at $S=0.53$, taking one large term $a_1\approx0.390908$ together with $10^5$ equal small terms summing to the remainder gives $P=0.53=S$ exactly (verified to 30 digits) — confirming the range is genuinely achievable, not merely necessary.

Summary Table

Property	Value
Lower bound on $P$	$P\ge1-S$ (Weierstrass)
Upper bound on $P$	$P
Range where $S=P$ is possible	$\tfrac12\le S<\Omega\approx0.567143$

Additional Notes

Refinement of the naive treatment: a draft analysis correctly identified $S<\Omega$ as necessary (from $P\Omega$, safely outside the achievable window, consistent with the mismatch $P\approx0.2888\ne1$ found by direct computation.
The omega constant $\Omega$ appearing here is the same constant defined by $\Omega e^\Omega=1$ (i.e. $\Omega=W(1)$ for the Lambert $W$ function), which also governs the asymptotic growth of iterated exponentials and appears in the analysis of the "power tower" $x^{x^{x^{\cdots}}}$.
The existence argument in Step 3 is a continuity/intermediate-value argument in disguise: as a fixed total mass $S$ is redistributed continuously from "one big term" (achieving $P=1-S$) to "infinitely many infinitesimal terms" (approaching $P\to e^{-S}$), $P$ sweeps through every value in $[1-S,e^{-S})$, so $S=P$ is achievable exactly when $S$ itself lies in that same interval.

Result

$$ \sum_{k=1}^{\infty}a_k=\prod_{k=1}^{\infty}(1-a_k)\ \text{ is possible for } 0