Question Number 161168 by mathocean1 last updated on 13/Dec/21 | ||
$${Show}\:{that}\:{U}_{{n}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{7}{n}+\mathrm{3}}\:\:{is}\:{convergent}\: \\ $$$${sequence}. \\ $$ | ||
Commented by MJS_new last updated on 13/Dec/21 | ||
$$\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{7}{n}+\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{7}}−\frac{\mathrm{19}}{\mathrm{49}{n}−\mathrm{21}} \\ $$ | ||
Answered by mathocean1 last updated on 14/Dec/21 | ||
$${thanks}... \\ $$ | ||
Answered by physicstutes last updated on 15/Dec/21 | ||
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{7}{n}+\mathrm{3}}\right)\:=\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\therefore\:\left\{{U}_{{n}} \right\}\:\mathrm{converges}\:\mathrm{to}\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$ | ||