This document presents a proof of convergence and boundedness of a sequence using induction, as part of a mathematics course or research project. It explores the properties of the sequence and its behavior over time, providing a detailed analysis of its convergence and boundedness.
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[...] We can conjecture that will be able to take the form of where are real On The last two equalities give us: so . We then get: . Finally: Let us show by induction that this formula is always true: - Initialisation : For , on a bien - Hérédité : suppose the formula is true at a certain rank . Then we have: Or : We finally get: The formula is therefore true at the rank - Conclusion : the formula being true at the rank and hereditary, it is true for all We deduce from the formula obtained that Exercise 2 is derivable on the interval of definition and we have: To build the first 4 terms of the sequence, we draw the line of equation Then we proceed as follows on the graph of the function provided by the statement: We can conjecture that the sequence is decreasing and converges to a value equal to 1. [...]
[...] is decreasing and . Thus is negative for . Or on a : car (demonstrated above) Thus then From previous inequalities, we deduce the sequel is decreasing and bounded, so it converges." On The sequel is a geometric progression of ratio and first term On so From where: According to question we have: Thus, tend vers then tend vers Exercice 3 The smallest of the terms is the following: The smallest term is the following: As there are terms, we deduce from the previous question: By stretching towards infinity, we deduce that the limit of the sequence est The algorithm above is looking for what value of the smallest possible, the sum S (which represents is strictly comprised between and . [...]
[...] Let us show by induction that we have: for any integer : - Initialisation : For , on a bien - Hérédité : suppose the formula is true at a certain rank . Then we have: Or, according to the first question, is growing. In addition . So for . We deduce that since (hypothesis of recurrence) As , on a et . We get Finally - Conclusion : the formula being true at the rank and hereditary, it is true for all It is deduced in particular that: Let us now . [...]
