Investigate the absolute and conditional convergence
Investigate the absolute and conditional convergence of $\displaystyle\prod_{n=1}^{\infty}\left(1+\frac{(-1)^{n+1}}n\right)$.
Explanation of Method
Since $(-1)^{n+1}=-(-1)^n$, this product is $\prod\bigl(1-(-1)^n/n\bigr)$ — exactly Problem 368's product, viewed through the opposite sign convention. Absolute convergence fails because $\sum1/n$ diverges, but the terms pair up into an odd/even chain that telescopes to exactly $1$, giving convergence in the conditional sense.
- Test absolute convergence via $\sum|a_n|=\sum1/n$.
- Pair consecutive terms and show each pair multiplies to $1$.
- Conclude conditional convergence and identify the limit.
Step 1 · Absolute convergence fails
With $a_n=(-1)^{n+1}/n$, $|a_n|=1/n$, and $\sum1/n$ is the divergent harmonic series, so $\prod(1+a_n)$ does not converge absolutely.
Step 2 · Pairing gives exact cancellation
For odd $n=2k-1$: factor is $1+\frac1{2k-1}=\frac{2k}{2k-1}$. For even $n=2k$: factor is $1-\frac1{2k}=\frac{2k-1}{2k}$. Each consecutive pair multiplies to exactly $1$: $$ \frac{2k}{2k-1}\cdot\frac{2k-1}{2k}=1. $$ So $P_{2M}=1$ for every $M$, and $P_{2M+1}=P_{2M}\cdot\frac{2M+2}{2M+1}=1+\frac1{2M+1}\to1$.
Step 3 · Conditional convergence to $1$
$$ \boxed{\prod_{n=1}^{\infty}\left(1+\frac{(-1)^{n+1}}n\right) = 1 \quad\text{(converges conditionally, not absolutely)}} $$ (Numerical check: $P_2=1$, $P_3=4/3$, $P_4=1$, $P_5=6/5$, $P_{100}=1$, $P_{101}=101/100$ — matching the two-subsequence pattern exactly.)
Summary Table
| Property | Value |
|---|---|
| Absolute convergence | fails ($\sum1/n=\infty$) |
| Conditional convergence | holds, to the value $1$ |
| $P_{2M}$ | exactly $1$ |
| $P_{2M+1}$ | $1+\dfrac1{2M+1}\to1$ |
Additional Notes
- This problem is mathematically identical to Problem 368: since $(-1)^{n+1}=-(-1)^n$, the factor $1+(-1)^{n+1}/n$ is the same as $1-(-1)^n/n$ term-for-term — the two problems simply present the alternating sign with opposite conventions, and both reduce to the identical telescoping pairing argument and the identical limit $1$.
- This is the sharpest possible illustration of "conditional" convergence for infinite products: the product converges via exact, finite cancellation in pairs, even though the corresponding series of magnitudes $\sum|a_n|$ diverges as badly as the harmonic series.
- Problem 380 generalizes this exact setup to $\prod(1+(-1)^{n+1}/n^\alpha)$, revealing that the phenomenon here ($\alpha=1$, conditional convergence to a nonzero value) sits in the middle of a three-way classification governed by two critical exponents, $\alpha=1$ (absolute) and $\alpha=1/2$ (any convergence at all).