U-n-is-a-sequence-wich-verify-lim-n-U-n-1-U-n-a-calculate-lim-n-U-n-n-

Question Number 64650 by mathmax by abdo last updated on 20/Jul/19

U_{n} i s a s e q u e n c e w i c h v e r i f y {l i m}_{n \to + \infty} U_{n + 1} - U_{n} = a

c a l c u l a t e {l i m}_{n \to + \infty} \frac{U_{n}}{n}

Commented by mathmax by abdo last updated on 20/Jul/19

\forall ξ > 0 \exists n_{o} i n t e g r / n > n_{0} \Rightarrow∣ u_{n + 1} - u_{n} - a ∣< ξ \Rightarrow

∣ \sum_{k = 0}^{n - 1} (u_{k + 1} - u_{k}) - n a ∣< n ξ \Rightarrow∣ u_{n} - u_{o} - n a ∣< n ξ \Rightarrow

∣ \frac{u_{n}}{n} - \frac{u_{o}}{n} - a ∣< ξ \Rightarrow {l i m}_{n \to + \infty} \frac{u_{n}}{n} = a .