Question Number 210549 by universe last updated on 12/Aug/24
$$\mathrm{let}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{be}\:\mathrm{difined}\:\mathrm{as} \\ $$$$\:\mathrm{a}_{\mathrm{n}} =\:\mathrm{a}_{\mathrm{n}−\mathrm{1}} \:+\:\frac{\mathrm{2cos}\:\left(\frac{\mathrm{a}_{\mathrm{n}−\mathrm{1}} }{\mathrm{2}}\right)}{\mathrm{2sin}\:\left(\frac{\mathrm{a}_{\mathrm{n}−\mathrm{1}} }{\mathrm{2}}\right)−\mathrm{1}}\:\:,\:\mathrm{a}_{\mathrm{0}\:} =\:\mathrm{0} \\ $$$$\mathrm{find}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{a}_{\mathrm{n}\:} \:=\:? \\ $$
Answered by aleks041103 last updated on 16/Aug/24
Commented by aleks041103 last updated on 16/Aug/24
$${f}\left({x}\right)\:=\:{x}\:+\:\frac{\mathrm{2}{cos}\left({x}/\mathrm{2}\right)}{\mathrm{2}{sin}\left({x}/\mathrm{2}\right)−\mathrm{1}} \\ $$$$ \\ $$$${a}_{{n}} ={f}\left({a}_{{n}−\mathrm{1}} \right) \\ $$$$ \\ $$$${seems}\:{like}\:{a}_{{n}} \rightarrow−\pi \\ $$$${the}\:{proof}\:{seems}\:{to}\:{be}\:{a}\:{bit}\:{difficult} \\ $$