Question Number 119750 by 675480065 last updated on 26/Oct/20
![Π_(k=1) ^(1019) [((2k)/(2k−1))]=?](Q119750.png)
$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{1019}} {\prod}}\left[\frac{\mathrm{2k}}{\mathrm{2k}−\mathrm{1}}\right]=? \\ $$
Answered by Bird last updated on 26/Oct/20
![Π_(k=1) ^(1019) [((2k)/(2k−1))] =Π_(k=1) ^(1019) [((2k−1+1)/(2k−1))] =Π_(k=1) ^(1019) [1+(1/(2k−1))] =Π_(k=1) ^(1019) (1+[(1/(2k−1))]) =2Π_(k=2) ^(1019) (1+[(1/(2k−1))]) but 0<(1/(2k−1))<1 for k∈[[2,1019]] ⇒ [(1/(2k−1))]=0 ⇒Π_(k=1) ^(1019) [((2k)/(2k−1))]=2](Q119751.png)
$$\prod_{{k}=\mathrm{1}} ^{\mathrm{1019}} \left[\frac{\mathrm{2}{k}}{\mathrm{2}{k}−\mathrm{1}}\right]\:=\prod_{{k}=\mathrm{1}} ^{\mathrm{1019}} \left[\frac{\mathrm{2}{k}−\mathrm{1}+\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right] \\ $$$$=\prod_{{k}=\mathrm{1}} ^{\mathrm{1019}} \left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right] \\ $$$$=\prod_{{k}=\mathrm{1}} ^{\mathrm{1019}} \left(\mathrm{1}+\left[\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right]\right) \\ $$$$=\mathrm{2}\prod_{{k}=\mathrm{2}} ^{\mathrm{1019}} \left(\mathrm{1}+\left[\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right]\right)\:\:{but} \\ $$$$\mathrm{0}<\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}<\mathrm{1}\:{for}\:{k}\in\left[\left[\mathrm{2},\mathrm{1019}\right]\right]\:\Rightarrow \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right]=\mathrm{0}\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{\mathrm{1019}} \left[\frac{\mathrm{2}{k}}{\mathrm{2}{k}−\mathrm{1}}\right]=\mathrm{2} \\ $$$$ \\ $$