Prove that product from n=2 to infinity (1+
Prove that $\displaystyle\prod_{n=2}^{\infty}\left(1+\frac1{2^n-2}\right)=\sum_{n=0}^{\infty}\frac1{2^n}=2$.
Explanation of Method
Each factor simplifies to $\frac{2^n-1}{2^n-2}=\frac{2^n-1}{2(2^{n-1}-1)}$, so writing $a_k=2^k-1$ turns the partial product into a telescoping ratio $a_N/a_1$ times an explicit power of $2$ — collapsing to a clean closed form that is exactly the $N$-th partial sum of the geometric series on the right.
- Simplify each factor and isolate a telescoping ratio $a_n/a_{n-1}$ with $a_k=2^k-1$.
- Telescope to get a closed form for the partial product $P_N$.
- Take the limit and match against the geometric series' partial sums.
Step 1 · Simplify each factor
$$ 1+\frac1{2^n-2}=\frac{2^n-1}{2^n-2}=\frac{2^n-1}{2(2^{n-1}-1)}. $$ With $a_k=2^k-1$: $$ P_N=\prod_{n=2}^N\left(1+\frac1{2^n-2}\right)=\frac1{2^{N-1}}\prod_{n=2}^N\frac{a_n}{a_{n-1}}. $$
Step 2 · Telescope
$$ \prod_{n=2}^N\frac{a_n}{a_{n-1}}=\frac{a_N}{a_1}=\frac{2^N-1}{1}=2^N-1 \quad(\text{since }a_1=2^1-1=1). $$ So $$ P_N=\frac{2^N-1}{2^{N-1}}=2-\frac1{2^{N-1}}. $$
Step 3 · Limit and identification with the geometric series
$$ \boxed{\lim_{N\to\infty}P_N=\lim_{N\to\infty}\left(2-\frac1{2^{N-1}}\right)=2} $$ In fact $P_N=1+\frac12+\frac14+\cdots+\frac1{2^{N-1}}$ is exactly the $N$-th partial sum of $\sum_{n=0}^\infty1/2^n$, so the identity holds term-by-partial-sum, not just in the limit. (Numerical check: $P_2=3/2$, $P_3=7/4$, $P_5=31/16=1.9375$, $P_{10}=1023/512\approx1.998$ — matching $2-2^{1-N}$ exactly at every step.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=2-\dfrac1{2^{N-1}}$ |
| Limit | $2$ |
| Matches | partial sum of $\sum_{n=0}^\infty2^{-n}$ |
Additional Notes
- The identity $P_N=2-2^{1-N}$ is not merely asymptotic — it exactly equals $\sum_{k=0}^{N-1}2^{-k}$ at every finite $N$, giving a genuinely exact correspondence between a finite product and a finite geometric sum, not just matching limits.
- This is philosophically the same phenomenon as Problem 366, where an infinite product of terms built from powers of $2$ collapses (via a telescoping identity) into a simple rational/geometric closed form.
- The convergence is exponentially fast, since the error term $2^{1-N}$ halves with each additional factor — contrast with Problem 369's polynomial ($O(1/N)$) rate of convergence for a telescoping rational product.