Prove convergence and find the value of product from n=1
Prove convergence and find the value of $\displaystyle\prod_{n=1}^{\infty}\left[1-(-1)^n\frac1n\right]$.
Explanation of Method
Splitting by parity, the odd-indexed factors become $\frac{2k}{2k-1}$ and the even-indexed factors become $\frac{2k-1}{2k}$ — reciprocals of each other. Pairing each odd term with the following even term makes every pair collapse to exactly $1$, so the partial products are either exactly $1$ (even $N$) or a single leftover factor close to $1$ (odd $N$), giving a limit of $1$.
- Split by parity into two interleaved sub-products.
- Pair consecutive odd/even terms and observe each pair multiplies to $1$.
- Take the limit, treating even and odd partial products separately.
Step 1 · Split by parity
For $n=2k-1$ (odd), $(-1)^n=-1$, so the factor is $1+\frac1{2k-1}=\frac{2k}{2k-1}$. For $n=2k$ (even), $(-1)^n=1$, so the factor is $1-\frac1{2k}=\frac{2k-1}{2k}$. So the product interleaves reciprocal pairs: $$ \left(\frac21\right)\left(\frac12\right)\left(\frac43\right)\left(\frac34\right)\left(\frac65\right)\left(\frac56\right)\cdots $$
Step 2 · Pair consecutive terms
Each consecutive odd/even pair multiplies to exactly $1$: $$ \frac{2k}{2k-1}\cdot\frac{2k-1}{2k}=1. $$ So for even $N=2M$: $$ P_{2M}=\prod_{k=1}^M\left(\frac{2k}{2k-1}\cdot\frac{2k-1}{2k}\right)=1. $$ For odd $N=2M+1$, one extra unpaired factor remains: $$ P_{2M+1}=P_{2M}\cdot\frac{2M+2}{2M+1}=1+\frac1{2M+1}. $$
Step 3 · Take the limit
$$ \lim_{M\to\infty}P_{2M}=1, \qquad \lim_{M\to\infty}P_{2M+1}=\lim_{M\to\infty}\left(1+\frac1{2M+1}\right)=1. $$ Both subsequences converge to the same value, so $$ \boxed{\prod_{n=1}^{\infty}\left[1-(-1)^n\frac1n\right]=1} $$ Convergence also follows from $\sum\ln|1-(-1)^n/n|\approx\sum(-1)^{n+1}/n-\frac1{2n^2}$, both of which converge (alternating harmonic series and a $p=2$ series). (Numerical check: $P_2=1$, $P_3=4/3$, $P_4=1$, $P_5=6/5$, $P_{100}=1$, $P_{101}=101/100=1.01$ — matching the two-subsequence pattern exactly.)
Summary Table
| Property | Value |
|---|---|
| $P_{2M}$ (even partial product) | exactly $1$ |
| $P_{2M+1}$ (odd partial product) | $1+\dfrac1{2M+1}\to1$ |
| Limit | $1$ |
Additional Notes
- This is the cleanest possible convergent alternating product: pairing terms gives literally exact cancellation to $1$ at every even step, rather than merely converging asymptotically — a stronger statement than convergence alone.
- The convergence criterion — $\sum(-1)^{n+1}/n$ converges to $\ln2$ (conditionally) while $\sum1/n^2$ converges absolutely — mirrors the general principle used in Problem 370 (where a diverging log-series signals a product converging to $0$) and Problem 369 (where an absolutely convergent log-series signals a nonzero limit).
- Contrast with Problem 367's superficially similar alternating product $\prod(1+(-1)^n/a^n)$: there the terms decay geometrically rather than like $1/n$, so no exact pairwise cancellation occurs, and the limit has no elementary closed form.