Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\left(1+\frac1{n^\alpha}\right)$.
Explanation of Method
For positive terms $1+a_n$ with $a_n\ge0$, the product converges to a positive finite limit exactly when $\sum a_n$ converges, and diverges to $+\infty$ otherwise — the infinite-product analogue of the comparison test. Here $a_n=1/n^\alpha$ is a $p$-series, so the whole question reduces to the classical $p$-series threshold at $\alpha=1$.
- State the product-convergence criterion in terms of $\sum a_n$.
- Apply the $p$-series test to $a_n=n^{-\alpha}$.
- Handle the boundary case $\alpha=1$ directly and the remaining ranges by comparison.
Step 1 · The convergence criterion
For $a_n\ge0$, $\prod_{n=1}^\infty(1+a_n)$ converges to a positive finite limit iff $\sum_{n=1}^\infty a_n$ converges (since $\ln(1+a_n)\sim a_n$ termwise, and both series have nonnegative terms). Here $a_n=n^{-\alpha}$, so convergence hinges on the $p$-series $\sum n^{-\alpha}$.
Step 2 · Case $\alpha>1$
The $p$-series $\sum n^{-\alpha}$ converges, so $\sum\ln(1+n^{-\alpha})$ converges (by comparison, since $\ln(1+n^{-\alpha})\sim n^{-\alpha}$), and the product converges to a positive finite limit.
Step 3 · Cases $\alpha\le1$
At $\alpha=1$: $\prod_{n=1}^N\left(1+\frac1n\right)=\prod_{n=1}^N\frac{n+1}n=N+1\to\infty$ (elementary telescoping), so the product diverges. For $0<\alpha<1$: $n^{-\alpha}>n^{-1}$, so $\sum n^{-\alpha}>\sum n^{-1}=\infty$, and the product diverges. For $\alpha\le0$: $1+n^{-\alpha}\ge2$ for all $n$, so $P_N\ge2^N\to\infty$.
$$ \boxed{\prod_{n=1}^{\infty}\left(1+\frac1{n^\alpha}\right)\ \text{converges to a positive finite limit iff }\alpha>1;\ \text{diverges to }+\infty\text{ otherwise.}} $$ (Numerical check: at $\alpha=2$, $P_{10}\approx3.68$, $P_{1000}\approx3.68$ — clearly bounded; at $\alpha=1$, $P_N=N+1$ exactly; at $\alpha=1/2$, $P_{10}\approx28.6$ and growing.)
Summary Table
| Range of $\alpha$ | Behavior |
|---|---|
| $\alpha>1$ | converges to a positive finite limit |
| $\alpha=1$ | diverges: $P_N=N+1$ |
| $\alpha\le1$ | diverges to $+\infty$ |
Additional Notes
- This is the master template for the entire batch of convergence problems in this range (372–380): every subsequent problem reduces, after algebraic manipulation, to comparing some series $\sum a_n$ against a $p$-series or an alternating series.
- The $\alpha=1$ boundary case is genuinely special: the telescoping identity $\prod_{n=1}^N(1+1/n)=N+1$ gives an exact, elementary rate of divergence (linear in $N$), unlike the merely qualitative log-divergence used for $0<\alpha<1$.
- Problem 372's product $\prod4n^2/(4n^2-1)=\prod(1+1/(4n^2-1))$ is a convergent special case of this same family (with effective $a_n\sim1/(4n^2)$, i.e. "$\alpha=2$"-type decay), and turns out to equal the classical Wallis constant $\pi/2$.