Prove convergence and find the value of product from n=0
Prove convergence and find the value of $\displaystyle\prod_{n=0}^{\infty}\frac{(2n+1)(2n+7)}{(2n+3)(2n+5)}$.
Explanation of Method
Each factor contributes two odd numbers to the numerator and two to the denominator; writing out the partial product as a product of two separately-telescoping ratios of consecutive odd-number chains reduces everything to a simple rational function of $N$.
- Split each factor's numerator and denominator into two interleaved chains of odd numbers.
- Telescope each chain separately.
- Combine and take the limit $N\to\infty$.
Step 1 · Identify the two telescoping chains
Writing the numerator's factors as $(2n+1)$ and $(2n+7)$, and the denominator's as $(2n+3)$ and $(2n+5)$, the partial product splits into $$ P_N=\prod_{n=0}^N\frac{(2n+1)(2n+7)}{(2n+3)(2n+5)} = \left(\prod_{n=0}^N\frac{2n+1}{2n+3}\right)\left(\prod_{n=0}^N\frac{2n+7}{2n+5}\right). $$
Step 2 · Telescope each chain
$$ \prod_{n=0}^N\frac{2n+1}{2n+3}=\frac13\cdot\frac35\cdot\frac57\cdots\frac{2N+1}{2N+3}=\frac1{2N+3}, $$ $$ \prod_{n=0}^N\frac{2n+7}{2n+5}=\frac75\cdot\frac97\cdot\frac{11}9\cdots\frac{2N+7}{2N+5}=\frac{2N+7}5. $$
Step 3 · Combine and take the limit
$$ P_N=\frac1{2N+3}\cdot\frac{2N+7}5=\frac{2N+7}{5(2N+3)}. $$ $$ \boxed{\lim_{N\to\infty}P_N=\lim_{N\to\infty}\frac{2+7/N}{10+15/N}=\frac15} $$ Convergence: $a_n=1-\dfrac8{4n^2+16n+15}\sim1-\dfrac2{n^2}$, and $\sum1/n^2$ converges, so $\sum\ln a_n$ converges absolutely and the product converges to a nonzero limit. (Numerical check: $P_0=7/15\approx0.467$, $P_2=11/35\approx0.314$, $P_{10}=27/115\approx0.235$, $P_{100}\approx0.2039$; direct product to $10^6$ terms gives $0.200002$.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{2N+7}{5(2N+3)}$ |
| Limit | $\dfrac15$ |
Additional Notes
- The general pattern — a ratio of two linear factors offset from two other linear factors by a fixed gap — always telescopes into two independent chains of arithmetic progressions, exactly as in Problems 360–362 but here with odd numbers instead of consecutive integers.
- The gap of $6$ between $(2n+1)$ and $(2n+7)$ (equivalently, $3$ steps of $2$) is what fixes the final limit at $1/5$: in general $\prod_{n=0}^\infty\frac{(2n+1)(2n+1+2m)}{(2n+3)(2n+3+2m-2)}$-type products limit to a ratio determined by the endpoints of the telescoping chains, here $1/(2\cdot3-1)=1/5$.
- Since $a_n\to1$ like $O(1/n^2)$ (absolutely summable), this sits with Problem 362 in the "converges to a nonzero limit" category, contrasting with Problem 370's product of $1/n^2$ (which diverges to $0$ because $\sum\ln(1/n^2)$ diverges to $-\infty$).