Prove that the product product from n=1 to infinity(1
Prove that the product $\displaystyle\prod_{n=1}^{\infty}(1\pm\alpha_n)$ (with $\alpha_n>0$) converges if and only if the series $\displaystyle\sum_{n=1}^{\infty}\alpha_n$ converges.
Explanation of Method
The logarithm turns the product into a series $\sum\ln(1\pm\alpha_n)$, and the elementary two-sided bound $\frac x{1+x}\le\ln(1+x)\le x$ (and its analogue for $\ln(1-x)$) sandwiches $\ln(1\pm\alpha_n)$ between constant multiples of $\alpha_n$ once $\alpha_n\to0$ — so the two series converge or diverge together by direct comparison.
- Forward direction: product converges $\implies\sum\ln(1+\alpha_n)$ converges $\implies\alpha_n\to0\implies\sum\alpha_n$ converges by comparison.
- Reverse direction: $\sum\alpha_n$ converges $\implies\alpha_n\to0\implies|\ln(1+\alpha_n)-\alpha_n|=O(\alpha_n^2)=o(\alpha_n)\implies\sum\ln(1+\alpha_n)$ converges.
- Repeat the argument for $(1-\alpha_n)$, using $\alpha_n<1$ eventually to keep factors positive.
Step 1 · Forward direction, for $\prod(1+\alpha_n)$
If $\prod(1+\alpha_n)$ converges to $P>0$, then $\sum\ln(1+\alpha_n)=\ln P<\infty$, so $\ln(1+\alpha_n)\to0$, hence $\alpha_n\to0$. For $\alpha_n<1$ (eventually true), the standard bound $$ \frac{\alpha_n}{1+\alpha_n}\le\ln(1+\alpha_n)\le\alpha_n $$ gives $\alpha_n\le2\ln(1+\alpha_n)$ once $\alpha_n<1$, and since $\sum\ln(1+\alpha_n)$ converges, comparison gives $\sum\alpha_n<\infty$.
Step 2 · Reverse direction, for $\prod(1+\alpha_n)$
If $\sum\alpha_n$ converges, then $\alpha_n\to0$, so eventually $\alpha_n<\tfrac12$. There, $$ |\ln(1+\alpha_n)-\alpha_n| = \left|\frac{\alpha_n^2}2-\frac{\alpha_n^3}3+\cdots\right| \le \frac{\alpha_n^2}{2(1-\alpha_n)} \le \alpha_n^2. $$ Since $\alpha_n\to0$, $\alpha_n^2=o(\alpha_n)$, so $\sum\alpha_n^2$ converges by comparison with $\sum\alpha_n$ (as $\alpha_n^2\le\alpha_n$ eventually), and therefore $\sum\ln(1+\alpha_n)=\sum\alpha_n-\sum(\alpha_n-\ln(1+\alpha_n))$ converges (difference of two convergent series). Hence $\prod(1+\alpha_n)=\exp\bigl(\sum\ln(1+\alpha_n)\bigr)$ converges to a positive finite limit.
Step 3 · The case $\prod(1-\alpha_n)$
Identically, using $|\ln(1-\alpha_n)|\le\dfrac{\alpha_n}{1-\alpha_n}\le2\alpha_n$ for $\alpha_n\le\tfrac12$: both directions transfer verbatim, giving $$ \boxed{\prod_{n=1}^{\infty}(1\pm\alpha_n)\ \text{converges to a positive finite limit} \iff \sum_{n=1}^{\infty}\alpha_n\ \text{converges}} $$ (with $\alpha_n<1$ eventually assumed for the $(1-\alpha_n)$ case, to keep every factor positive).
Summary Table
| Direction | Key step |
|---|---|
| $\prod\Rightarrow\sum$ | $\alpha_n\le2\ln(1+\alpha_n)$ once $\alpha_n<1$ |
| $\sum\Rightarrow\prod$ | $|\ln(1+\alpha_n)-\alpha_n|\le\alpha_n^2=o(\alpha_n)$ |
Additional Notes
- This theorem is the single most-used tool across Problems 371–380: every convergence classification in that batch (the $p$-series thresholds at $\alpha=1$, $\alpha=1/2$, the geometric threshold $|x|<3$, and so on) is a direct application of this criterion to a specific choice of $\alpha_n$.
- The hypothesis "constant sign" ($\alpha_n>0$ throughout) is essential: Problem 385 constructs an explicit counterexample showing the equivalence breaks down completely once the $\alpha_n$ are allowed to alternate in sign, even though the same algebraic machinery (logarithms, Taylor expansion) is used.
- The bound $\alpha_n-\alpha_n^2\le\ln(1+\alpha_n)\le\alpha_n$ used here is the same elementary inequality underlying Problem 383's analysis, where it appears in the guise of $-\ln(1-a_k)\ge a_k$ (Weierstrass's inequality is precisely its exponentiated, summed form).