Evaluate the sum

\begin{aligned} \exp{(S)} &= \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdots\\ &= \prod_{n=1}^\infty \frac{2n}{2n-1}\cdot \frac{2n}{2n+1} \\ &= \frac{\pi}{2} \tag{Wallis product} \end{aligned}

Hence, $\lfloor 10^6\times S \rfloor = \lfloor 10^6\times \ln\left(\frac{\pi}{2}\right) \rfloor = 451582$

Relevant wiki: Wallis product

By writing out the summation, we see that, $S=\sum_{n=1}^{\infty}(-1)^{n-1}\ln{\left(\frac{n+1}{n}\right)}=\ln{\left(2\right)}-\ln{\left(\frac{3}{2}\right)}+\ln{\left(\frac{4}{3}\right)}-\ln{\left(\frac{5}{4}\right)}+\ln{\left(\frac{6}{5}\right)}-\ln{\left(\frac{7}{6}\right)}+\ln{\left(\frac{8}{7}\right)}-\ln{\left(\frac{9}{8}\right)}+...$ Applying the logarithmic product and quotient rules: $\sum_{n=1}^{\infty}(-1)^{n-1}\ln{\left(\frac{n+1}{n}\right)}=\ln{\left(\frac{4}{3}\times\frac{16}{15}\times\frac{36}{35}\times\frac{64}{63}\times...\right)}=\ln{\left(\prod_{n=1}^{\infty}\frac{4n^{2}}{4n^{2}-1}\right)}=\ln{\left(\prod_{n=1}^{\infty}\frac{2n}{2n-1}\times\frac{2n}{2n+1}\right)}$ Evaluating the product gives $\frac{\pi}{2}$ (Note: It is actually the Wallis' product), so $\sum_{n=1}^{\infty}(-1)^{n-1}\ln{\left(\frac{n+1}{n}\right)}=\ln{\left(\frac{\pi}{2}\right)}$ $\therefore\lfloor10^{6}\times S\rfloor=\lfloor10^{6}\times\ln{\left(\frac{\pi}{2}\right)}\rfloor=\lfloor451582.705289\rfloor=\boxed{451582}$