Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\left[\ln\left(1+\frac xn\right)\right]^{1/n}$ for $x>0$.
Explanation of Method
For $x>0$ each factor $\ln(1+x/n)$ is a positive number less than $1$ for large $n$, so raising it to the $1/n$ power still leaves something close to (but not exceeding) $1$. Taking logarithms turns the product into $\sum\frac1n\ln[\ln(1+x/n)]$, and the dominant asymptotic behavior $\ln[\ln(1+x/n)]\sim\ln x-\ln n$ makes this sum behave like $\ln x\cdot\ln N-\tfrac12(\ln N)^2\to-\infty$.
- Establish the asymptotic form $\ln[\ln(1+x/n)]=\ln x-\ln n+o(1)$.
- Sum $\frac1n[\ln x-\ln n]$ using the known asymptotics $\sum_{n\le N}\frac{\ln n}n\sim\frac{(\ln N)^2}2$.
- Conclude $\ln P\to-\infty$, hence $P=0$.
Step 1 · Asymptotic form of the general term
For $x>0$, $\ln(1+x/n)\sim x/n\to0$, so $$ \ln\left[\ln\left(1+\frac xn\right)\right]=\ln\frac xn+o(1)=\ln x-\ln n+o(1). $$
Step 2 · Sum the logarithmic series
Using $\sum_{n=1}^N\frac1n\sim\ln N$ and the classical asymptotic $\sum_{n=1}^N\frac{\ln n}n\sim\frac{(\ln N)^2}2$: $$ \ln P_N=\sum_{n=1}^N\frac1n\ln\left[\ln\left(1+\frac xn\right)\right]\sim\ln x\cdot\ln N-\frac{(\ln N)^2}2 = \ln N\left(\ln x-\frac{\ln N}2\right). $$
Step 3 · The limit
Since $\ln N\to\infty$ while $\ln x-\frac{\ln N}2\to-\infty$, the product of the two factors $\to-\infty$: $$ \boxed{\ln P_N\to-\infty \implies \prod_{n=1}^{\infty}\left[\ln\left(1+\frac xn\right)\right]^{1/n}=0\ \text{ for every }x>0.} $$ (Numerical check at $x=1$: $P_{100}\approx1.36\times10^{-5}$, $P_{10000}\approx2.1\times10^{-19}$, $P_{100000}\approx9.2\times10^{-30}$ — decaying superpolynomially, confirming $P\to0$; similarly at $x=5$.)
Summary Table
| Property | Value |
|---|---|
| Domain | $x>0$ |
| Asymptotic term | $\ln[\ln(1+x/n)]\sim\ln x-\ln n$ |
| Limit | $0$, for every $x>0$ |
Additional Notes
- The mechanism here differs from a simple "terms less than $1$" argument: individual factors $[\ln(1+x/n)]^{1/n}$ actually tend to $1$ as $n\to\infty$ (since taking an $n$-th root of a quantity vanishing only like $1/n$ washes out the smallness). The product still collapses to $0$ because the accumulated logarithmic deficit $-\tfrac12(\ln N)^2$ grows without bound, even though no single factor is far from $1$.
- This is a instructive companion to Problem 370's $\prod1/n^2=0$: both illustrate "degenerate" convergence to zero, but here the mechanism is a slowly-accumulating double-logarithmic divergence rather than a a straightforward polynomial decay.
- The result holds for every fixed $x>0$, showing the divergence rate ($\sim-\tfrac12(\ln N)^2$) is universal and independent of $x$ at leading order — $x$ only shifts the sub-leading term $\ln x\cdot\ln N$.