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

$$Showthat{U}_n=\frac{4n-1}{7n+3}isconvergent$$$$sequence.$$

Commented by MJS_new last updated on 13/Dec/21

$$\frac{4n-1}{7n+3}=\frac{4}{7}-\frac{19}{49n-21}$$

Answered by mathocean1 last updated on 14/Dec/21

$$thanks...$$

Answered by physicstutes last updated on 15/Dec/21

$$\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{4n-1}{7n+3}\right)=\frac{4}{7}$$$$\therefore \left\{U_n\right\}\mathrm{converges}\mathrm{to}\frac{4}{7}$$

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