Investigate the convergence of product from n=2
Investigate the convergence of $\displaystyle\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{\ln n}\right)$.
Explanation of Method
All factors are positive (the only borderline case, $n=2$ with $\ln2<1$, still gives a positive factor $1+1/\ln2$). The series $\sum1/\ln n$ diverges (even more slowly than the harmonic series, since $\ln n=o(n^\epsilon)$ for any $\epsilon>0$), so there is no absolute convergence; but splitting $\ln(1+a_n)$ into an alternating linear part (convergent, by the alternating series test) and a fixed-sign quadratic part $\sim1/(2(\ln n)^2)$ — which itself diverges since $(\ln n)^2$ grows far slower than any positive power of $n$ — forces the product to $0$.
- Confirm all factors are positive and well-defined.
- Show absolute convergence fails via $\sum1/\ln n=\infty$.
- Split the logarithmic series into an alternating (convergent) part and a quadratic (divergent) part to conclude $P=0$.
Step 1 · All factors are positive
For $n\ge2$, $\ln n>0$, so $1+(-1)^n/\ln n>0$ requires checking only odd $n$ where the sign is negative: $1-1/\ln n>0\iff\ln n>1\iff n>e$. Since $n\ge3$ for odd terms beyond $n=2$ (which has $+$ sign), and $3>e$, every factor is strictly positive.
Step 2 · No absolute convergence
Since $\ln n$ grows slower than any positive power of $n$, $\sum1/\ln n$ diverges even more slowly than the harmonic series (confirmed by the integral test, $\int^\infty dx/\ln x=\infty$), so the product does not converge absolutely.
Step 3 · The quadratic term forces $P=0$
Expanding: $$ \ln\left(1+\frac{(-1)^n}{\ln n}\right)=\frac{(-1)^n}{\ln n}-\frac1{2(\ln n)^2}+O\!\left(\frac1{(\ln n)^3}\right). $$ The first (alternating) sum converges by the alternating series test. The second sum, $\sum1/(2(\ln n)^2)$, diverges to $+\infty$ (again since $(\ln n)^2$ grows far slower than $n$, so this fails the integral test even more badly than $\sum1/\ln n$). With an overall minus sign, this drives $$ \boxed{\sum_{n=2}^{\infty}\ln\left(1+\frac{(-1)^n}{\ln n}\right)=-\infty \implies \prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{\ln n}\right)=0.} $$ (Numerical check: $P_{10}\approx0.125$, $P_{100}\approx0.0037$, $P_{1000}\approx1.5\times10^{-8}$, $P_{10000}\approx1.8\times10^{-36}$ — decaying at an accelerating rate, confirming $P\to0$.)
Summary Table
| Property | Value |
|---|---|
| Absolute convergence | fails ($\sum1/\ln n=\infty$) |
| Limit | $0$ |
Additional Notes
- This is structurally the same phenomenon as Problem 380's $\alpha\le1/2$ regime: an alternating linear term that converges paired with a fixed-sign quadratic term that diverges, forcing the product to $0$ rather than merely failing absolute convergence.
- The extremely slow growth of $\ln n$ (slower than $n^\epsilon$ for any $\epsilon>0$) makes the divergence of $\sum1/(\ln n)^2$ especially severe compared to, say, Problem 377's $\sum n^{-2\alpha}$ — correspondingly, the product here decays to $0$ faster (in $N$) than any of the $\alpha\le1/2$ cases in that problem.
- Unlike Problem 368 or 379, where the alternating factors are $O(1/n)$ and pair up to telescope exactly to a finite nonzero value, here the much slower decay of $1/\ln n$ means the terms never approach $1$ fast enough for that kind of exact cancellation — the quadratic correction dominates instead.