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**Guys,**

I'm trying to prove by induction that the sequence given by[tex] a_{n+1}=3-\frac{1}{a_n} \qquad a_1=1 [/tex]

I'm trying to prove by induction that the sequence given by

**is increasing and**[tex] a_n < 3 \qquad \forall n .[/tex]

**Is the following correct? Thank you.**

*Task #1.*

[tex] n = 1 \Longrightarrow a_2=2>a_1 [/tex] is true.

We assume [tex] n = k [/tex] is true. Then,

[tex] 3-\frac{1}{a_{k+1}} > 3-\frac{1}{a_k} [/tex]

[tex] a_{k+2} > a_{k+1} [/tex] is true for [tex] n=k+1 [/tex].

This shows, by mathematical induction, that

[tex] a_{n+1} > a_{n} \qquad \forall n . [/tex]

*Task #2*

We already know that

[tex] a_1 < 3 [/tex] is true.

We assume [tex] n=k [/tex] is true. Then,

[tex] a_k < 3 [/tex]

[tex] \frac{1}{a_k} > \frac{1}{3} [/tex]

[tex] -\frac{1}{a_k} < -\frac{1}{3} [/tex]

[tex] 3-\frac{1}{a_k} < 3-\frac{1}{3} [/tex]

[tex] a_{k+1} < \frac{8}{3} < 3 [/tex]

[tex] a_{k+1} < 3 [/tex] is true for [tex] n = k+1 [/tex]. Thus,

[tex] a_{n} < 3 \qquad \forall n . [/tex]

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