\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \dots \cdot \frac{2k}{2k+1} \cdot \frac{2(k+1)}{2(k+1)+1}  &>&  \frac{1}{\sqrt{2k+1}} \cdot \frac{2(k+1)}{2(k+1)+1}\,\ \\ &=& \frac{1}{\sqrt{2k+1}} \cdot \frac { \sqrt{(2(k+1))^2}} { \sqrt{(2(k+1)+1)^2} }\\ &=& \frac{1}{\sqrt{2k+1}} \cdot \frac {\sqrt {4k^2+8k+4}}{\sqrt {2k+3} \cdot \sqrt {2(k+1)+1}} \,\  \\ &=& \frac {\sqrt {4k^2+8k+4}} { \sqrt{4k^2+8k+3}} \cdot \frac {1}{\sqrt{2(k+1)+1}}\,\ \\ &>& \,\ 1 \,\ \cdot \,\ \frac {1}{\sqrt{2(k+1)+1}}\\ &=& \frac {1}{\sqrt{2(k+1)+1}}