Can you use mathematical induction to prove that t_n >= t_(n-1) for all n in ZZ^+ for a sequence with the general term: t_n=(3n+5)/(n+2), n in ZZ^+?

(b) And hence, or otherwise, prove that 8/3 <= t_n <= 3 for all n in ZZ^+

1 Answer
Feb 7, 2017

Induction does not seem to help prove the initial conjecture, but seems better suited for proving part (b).


Proof: t_n-t_(n-1) = (3n+5)/(n+2) - (3(n-1)+5)/((n-1)+2)

=(3n+5)/(n+2)-(3n+2)/(n+1)

=((3n+5)(n+1)-(3n+2)(n+2))/((n+1)(n+2))

=1/((n+1)(n+2)

>0 for all n in ZZ^+

:. t_n > t_(n-1) for all n in ZZ^+


(b)

Proof: (by induction)

Base case: For n=1, we have t_1 = 8/3 in [8/3, 3].

Inductive hypothesis: Suppose that t_k in [8/3, 3] for some k in ZZ^+.

Induction step: We wish to show that t_(k+1) in [8/3, 3]. Indeed,

8/3 <= t_k" " (by the inductive hypothesis)

<= t_(k+1)" " (by the previous proof)

= (3n+8)/(n+3)

< (3n+9)/(n+3)

=3

:. t_(k+1) in [3/8, 3]

We have supposed true for k and shown true for k+1, thus, by induction, t_n in [3/8, 3] for all n in ZZ^+. ∎