let-u-0-5-and-n-N-u-n-1-u-n-1-n-prove-that-45-lt-u-1000-lt-45-1-

Tinku Tara June 4, 2023 Relation and Functions 0 Comments

Question Number 50392 by Abdo msup. last updated on 16/Dec/18

let u_0 =5 and ∀n∈N u_(n+1) =u_n +(1/n) prove that 45<u_(1000) <45,1

$${let}\:{u}_{\mathrm{0}} =\mathrm{5}\:{and}\:\forall{n}\in{N}\:\:\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} \:+\frac{\mathrm{1}}{{n}} \\ $$$${prove}\:{that}\:\mathrm{45}<{u}_{\mathrm{1000}} <\mathrm{45},\mathrm{1} \\ $$

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let-u-0-5-and-n-N-u-n-1-u-n-1-n-prove-that-45-lt-u-1000-lt-45-1-

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