Menu Close

Category: Relation and Functions

Question-145960

Question Number 145960 by puissant last updated on 10/Jul/21 Answered by mathmax by abdo last updated on 10/Jul/21 $$\mathrm{R}_{\mathrm{n}} =\sum_{\mathrm{p}=\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}} \mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{p}}\right)\:\Rightarrow\mathrm{R}_{\mathrm{n}} =_{\mathrm{p}−\mathrm{n}=\mathrm{k}} \:\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}}…