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^+#?

Question

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

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Answer 1 · 2017-02-07T11:09:28

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^+#. ∎