Limit $\lim_{k\to\infty}\prod_{m=2^k}^{m=2^{k+1}} {\frac{2m}{2m-1}}$

Question

Does $$\lim_{k\to\infty}\prod_{m=2^k}^{m=2^{k+1}} {\frac{2m}{2m-1}}$$ exist? Wolfram alpha gives numbers around 1.4

Answer 1

Define $$ \begin{align*} p_k&=\prod_{m=2^k}^{2^{k+1}}\frac{2m}{2m-1}\\ \log p_k&=\sum_{2^k}^{2^{k+1}}\log\left(1+\frac{1}{2m-1}\right)\\ &=\sum_{2^k}^{2^{k+1}}\left[\frac{1}{2m-1}+\frac{O(1)}{(2m-1)^2}\right]\\ &=\int_{2^k}^{2^{k+1}}\frac{1}{2x}\ dx+O(2^{-k})\\ &=\frac{\log (2^{k+1})-\log(2^k)}{2}+O(2^{-k})\\ &=\frac{\log 2}{2}+O(2^{-k}). \end{align*}$$ Consequently the limit as $k\to\infty$ exists and equals $\sqrt{2}$.

Answer 2

Let $$A_k=\prod_{m=2^k}^{m=2^{k+1}}\frac{2m}{2m-1}\\ B_k=\prod_{m=2^k}^{m=2^{k+1}-1}\frac{2m+1}{2m}\\ A_kB_k=2^{k+2}/(2^{k+1}-1)\\ \frac{2^{k+2}+1}{2^{k+2}}B_k<A_k<\frac{2^{k+1}-1}{2^{k+1}-2}B_k$$ $A_k/B_k\to1$, and $A_kB_k\to2$
$\lim_{k\to\infty}A_k^2=\lim_{k\to\infty}(A_k/B_k)\lim_{k\to\infty}A_kB_k=1\times2$