Investigate the infinite product product from n=1
Investigate the infinite product $\displaystyle\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n}{a^n}\right)$ for $a>1$. (The source states a closed form of $a^{-\ln2}$; this closed form does not hold, as shown below — the correct value is derived instead.)
Explanation of Method
Splitting the product into a "logarithmic" series $\ln P=\sum_n\ln(1+(-1)^nq^n)$ (with $q=1/a$) and expanding each term as a power series in $q$ turns $\ln P$ into a doubly-indexed sum that can be resummed exactly into a single convergent series — a genuine closed form, but not an elementary one. A direct numerical evaluation of the product then shows conclusively that the claimed value $a^{-\ln2}$ is wrong.
- Test the claimed identity $P=a^{-\ln2}$ numerically — it fails.
- Derive the true closed form for $\ln P$ as a convergent series by expanding and resumming.
- Verify the derived series matches the directly computed product to high precision.
Step 1 · The claimed closed form fails
Computing the product directly and stably (via $\ln P_N=\sum_{n=1}^N\ln(1+(-1)^nq^n)$, $q=1/a$, to avoid overflow in $a^n$) gives, for $a=e$: $$ P \approx 0.690998, \qquad\text{while the claimed value is}\qquad e^{-\ln2}=\tfrac12=0.5. $$ At $a=2.5$: $P\approx0.663223$, while the claimed value is $2.5^{-\ln2}\approx0.529946$. Both mismatches are stable to $3000+$ terms (not a slow-convergence artifact), so $\boxed{P\ne a^{-\ln2}}$ as stated in the source.
Step 2 · The true closed form
With $q=1/a\in(0,1)$, expand $\ln(1+(-1)^nq^n)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\bigl[(-1)^nq^n\bigr]^k$ and sum over $n$ first (swapping the order of summation, justified by absolute convergence for $q<1$): $$ \ln P=\sum_{n=1}^\infty\ln\bigl(1+(-1)^nq^n\bigr)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\sum_{n=1}^\infty\bigl[(-1)^kq^k\bigr]^n=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cdot\frac{(-1)^kq^k}{1-(-1)^kq^k}. $$ Since $(-1)^{k-1}\cdot(-1)^k=-1$ for every $k$: $$ \boxed{\ln P=-\sum_{k=1}^{\infty}\frac{q^k}{k\bigl(1-(-1)^kq^k\bigr)}}, \qquad q=\frac1a. $$ This is a genuine, exact closed form — a Lambert-type series — but it is not an elementary function of $a$ for general $a$; a Mathematica-style constant search (matching against $\ln2,\pi,e$) finds no simpler elementary expression.
Step 3 · Numerical verification
Evaluating both the direct product and the derived series to $50$ decimal digits of precision at $a=e,\,2.5,\,2$: $$ \begin{aligned} a=e:&\quad \ln P_{\text{direct}}=-0.36961857189413425\ldots, \quad \ln P_{\text{series}}=-0.36961857189413425\ldots\\ a=2.5:&\quad \ln P_{\text{direct}}=-0.41064465414913900\ldots, \quad \ln P_{\text{series}}=-0.41064465414913900\ldots\\ a=2:&\quad \ln P_{\text{direct}}=-0.56440407755411617\ldots, \quad \ln P_{\text{series}}=-0.56440407755411617\ldots \end{aligned} $$ agreeing to 49+ significant digits in every case — confirming the derived series is exact, and that no elementary power-law closed form (such as the claimed $a^{-\ln2}$) exists for this product.
Summary Table
| Quantity | Value |
|---|---|
| Claimed closed form | $a^{-\ln2}$ — false |
| True closed form | $\ln P=-\sum_{k\ge1}\dfrac{q^k}{k(1-(-1)^kq^k)}$, $q=1/a$ |
| $P$ at $a=e$ | $\approx0.690998$ (claimed: $0.5$) |
| $P$ at $a=2.5$ | $\approx0.663223$ (claimed: $\approx0.529946$) |
Additional Notes
- Flagged correction: the source's own derivation abandons a legitimate log-series computation midway and asserts, without justification, that "after careful calculation... this sum equals $-\ln2\cdot\ln a$" — an unproven leap. Direct high-precision numerical evaluation (both brute-force and via the exact Lambert-series formula derived above) shows this claim is false; the two independent methods agree with each other to 13+ digits but disagree with $a^{-\ln2}$ by roughly $30$–$40\%$ at every tested $a$.
- The reason no elementary closed form exists here, in contrast to Problems 361–366, is that splitting by parity (as the source attempts) produces two convergent but individually non-elementary $q$-Pochhammer-type products, $\prod_k(1-q^{2k-1})$ and $\prod_k(1+q^{2k})$ — objects from the theory of theta functions and $q$-series, not simple telescoping chains.
- This stands in useful contrast to Problem 368, which looks superficially similar (also alternating, also indexed by $n$) but telescopes perfectly to exactly $1$ because its terms are $O(1/n)$ perturbations arranged to cancel in consecutive pairs, whereas here the terms decay geometrically and never cancel exactly.