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Question Number 161168 by mathocean1 last updated on 13/Dec/21

Show that U_n =((4n−1)/(7n+3))  is convergent   sequence.

$${Show}\:{that}\:{U}_{{n}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{7}{n}+\mathrm{3}}\:\:{is}\:{convergent}\: \\ $$$${sequence}. \\ $$

Commented by MJS_new last updated on 13/Dec/21

((4n−1)/(7n+3))=(4/7)−((19)/(49n−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...

$${thanks}... \\ $$

Answered by physicstutes last updated on 15/Dec/21

lim_(n→∞) (((4n−1)/(7n+3))) = (4/7)  ∴ {U_n } converges to (4/7)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{7}{n}+\mathrm{3}}\right)\:=\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\therefore\:\left\{{U}_{{n}} \right\}\:\mathrm{converges}\:\mathrm{to}\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$

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