Investigate the convergence of the infinite product
Investigate the convergence of the infinite product $\displaystyle\prod_{n=1}^{\infty}\frac1{n^2}$.
Explanation of Method
An infinite product $\prod a_n$ (with $a_n>0$) converges to a nonzero limit exactly when $\sum\ln a_n$ converges; here $\ln a_n=-2\ln n\to-\infty$ so quickly that the logarithmic series diverges to $-\infty$, forcing the product itself to $0$ — a degenerate but well-defined form of convergence.
- Apply the log-series convergence criterion to $a_n=1/n^2$.
- Show $\sum\ln n$ diverges to $+\infty$ via the integral test.
- Conclude and cross-check with the exact closed form $P_N=1/(N!)^2$.
Step 1 · The log-series criterion
With $a_n=1/n^2$: $\ln a_n=-2\ln n$, so $$ \sum_{n=1}^\infty\ln a_n=-2\sum_{n=1}^\infty\ln n. $$
Step 2 · $\sum\ln n$ diverges
By the integral test, $\displaystyle\int_2^N\ln x\,dx=[x\ln x-x]_2^N=N\ln N-N-2\ln2+2\to+\infty$ as $N\to\infty$, so $\sum_{n=1}^\infty\ln n=+\infty$, hence $\sum_{n=1}^\infty\ln a_n=-\infty$.
Step 3 · Conclusion and direct check
Since the log-series diverges to $-\infty$, the product converges to $0$: $$ \boxed{\prod_{n=1}^{\infty}\frac1{n^2}=0} $$ Directly, $P_N=\dfrac1{1^2\cdot2^2\cdots N^2}=\dfrac1{(N!)^2}$, and by Stirling's approximation $(N!)^2\sim2\pi N(N/e)^{2N}\to\infty$ super-exponentially, so $P_N\to0$. (Numerical check: $P_5=1/14400\approx6.94\times10^{-5}$, $P_{10}=1/(10!)^2\approx7.6\times10^{-14}$.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac1{(N!)^2}$ |
| Limit | $0$ |
| Log-series behavior | diverges to $-\infty$ |
Additional Notes
- By convention, an infinite product is usually said to diverge to $0$ (rather than "converge to $0$") whenever infinitely many factors push the partial products to $0$ without any single factor being exactly $0$ — this problem is the standard textbook illustration of that degenerate case, distinct from genuine convergence to a nonzero limit.
- This is the polar opposite of Problem 369, where $a_n\to1$ like $1-2/n^2$ and the absolutely convergent log-series gives a nonzero limit ($1/5$); here $a_n=1/n^2\to0$ itself, so no amount of telescoping can rescue a nonzero limit.
- The factorial-squared closed form $P_N=1/(N!)^2$ decays faster than any exponential $r^N$ (for any fixed $r$), making this the fastest-decaying product encountered in this collection — Stirling's formula quantifies the decay rate precisely as $\sim(e/N)^{2N}/(2\pi N)$.