Show that Problem 384's criterion fails once the alpha_n
Show that Problem 384's criterion fails once the $\alpha_n$ do not have constant sign, by verifying that $$ \left(1+\frac1{\sqrt2}+\frac12\right)\left(1-\frac1{\sqrt2}\right)\left(1+\frac1{\sqrt3}+\frac13\right)\left(1-\frac1{\sqrt3}\right)\cdots $$ converges although the corresponding series $\left(\frac1{\sqrt2}+\frac12\right)+\left(-\frac1{\sqrt2}\right)+\left(\frac1{\sqrt3}+\frac13\right)+\left(-\frac1{\sqrt3}\right)+\cdots$ diverges.
Explanation of Method
Multiplying out each consecutive pair of factors reveals that the $1/\sqrt n$ terms cancel completely, leaving $1-1/n^{3/2}$: a genuinely convergent product (by Problem 384, applied honestly to the positive quantity $1/n^{3/2}$). Meanwhile the underlying series, read term by term without pairing, is exactly $\sum1/n$ — the divergent harmonic series. The two facts coexist because the series' terms alternate in sign, violating Problem 384's hypothesis.
- Expand each paired factor $\bigl(1+\tfrac1{\sqrt n}+\tfrac1n\bigr)\bigl(1-\tfrac1{\sqrt n}\bigr)$ algebraically.
- Identify the resulting product as $\prod(1-1/n^{3/2})$, convergent since $\sum1/n^{3/2}$ converges.
- Identify the corresponding (unpaired) series as $\sum1/n$, the divergent harmonic series.
Step 1 · The paired product collapses
$$ \left(1+\frac1{\sqrt n}+\frac1n\right)\left(1-\frac1{\sqrt n}\right) = 1-\frac1{\sqrt n}+\frac1{\sqrt n}-\frac1n+\frac1n-\frac1{n^{3/2}} = 1-\frac1{n^{3/2}}. $$ (Numerical check at $n=2$: $(1+0.7071+0.5)(1-0.7071)=2.2071\times0.2929\approx0.6464$, matching $1-2^{-3/2}\approx0.6464$; at $n=3$: $1.9107\times0.4226\approx0.8074$, matching $1-3^{-3/2}\approx0.8075$.)
Step 2 · The product converges
So the given infinite product, grouped in its natural pairs, is $$ \prod_{n=2}^{\infty}\left(1-\frac1{n^{3/2}}\right), $$ which converges to a positive finite limit by Problem 384, since $\sum1/n^{3/2}$ is a convergent $p$-series ($p=3/2>1$).
Step 3 · The series diverges
Reading off the terms of the corresponding series exactly as they are listed (not grouped by pairs): each pair contributes $$ \left(\frac1{\sqrt n}+\frac1n\right)+\left(-\frac1{\sqrt n}\right)=\frac1n, $$ so the full series is $$ \sum_{n=2}^{\infty}\frac1n = \infty \qquad\text{(the divergent harmonic series).} $$ $$ \boxed{\text{Product converges}\ (\text{to } \textstyle\prod(1-n^{-3/2})),\qquad \text{Series diverges}\ (\textstyle\sum1/n=\infty).} $$
Summary Table
| Object | Behavior |
|---|---|
| Product (paired) | converges: $\prod(1-n^{-3/2})$ |
| Series (unpaired terms) | diverges: $\sum1/n=\infty$ |
Additional Notes
- The mechanism is the same "cancellation via pairing" seen in Problems 368 and 379, but pushed one step further: there the pairing gave an exact factor of $1$ (product value unaffected by the pairing choice); here the pairing produces a genuinely smaller, convergent quantity $1-n^{-3/2}$ out of two individually large-ish alternating terms $\pm1/\sqrt n$.
- This example shows precisely why Problem 384's proof breaks down for alternating $\alpha_n$: the key step there used $\alpha_n\to0$ to control $|\ln(1+\alpha_n)-\alpha_n|=O(\alpha_n^2)$, but with terms as large as $1/\sqrt n$ individually, no such control exists termwise — only the algebraic cancellation between adjacent terms rescues convergence.
- The particular exponent $3/2$ emerging from the cancellation is not accidental: choosing the "linear" part of the first factor to be $1/\sqrt n$ (matching the second factor's magnitude) is exactly what makes the leading-order term cancel, leaving a residual of one order higher, $n^{-1}\cdot n^{-1/2}=n^{-3/2}$.