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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  is a sequence wich verify lim_(n→+∞) U_(n+1) −U_n =a     calculate lim_(n→+∞)  (U_n /n)
$${U}_{{n}} \:{is}\:{a}\:{sequence}\:{wich}\:{verify}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}+\mathrm{1}} −{U}_{{n}} ={a}\:\:\: \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\frac{{U}_{{n}} }{{n}} \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
∀ξ>0  ∃n_o  integr / n>n_0  ⇒∣u_(n+1) −u_n −a∣<ξ ⇒  ∣Σ_(k=0) ^(n−1) (u_(k+1) −u_k )−na∣<nξ ⇒∣u_n −u_o  −na∣<nξ ⇒  ∣(u_n /n) −(u_o /n) −a∣<ξ ⇒  lim_(n→+∞)   (u_n /n) =a .
$$\forall\xi>\mathrm{0}\:\:\exists{n}_{{o}} \:{integr}\:/\:{n}>{n}_{\mathrm{0}} \:\Rightarrow\mid{u}_{{n}+\mathrm{1}} −{u}_{{n}} −{a}\mid<\xi\:\Rightarrow \\ $$$$\mid\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({u}_{{k}+\mathrm{1}} −{u}_{{k}} \right)−{na}\mid<{n}\xi\:\Rightarrow\mid{u}_{{n}} −{u}_{{o}} \:−{na}\mid<{n}\xi\:\Rightarrow \\ $$$$\mid\frac{{u}_{{n}} }{{n}}\:−\frac{{u}_{{o}} }{{n}}\:−{a}\mid<\xi\:\Rightarrow\:\:{lim}_{{n}\rightarrow+\infty} \:\:\frac{{u}_{{n}} }{{n}}\:={a}\:. \\ $$

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