# 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 $\frac{8}{3} \le {t}_{n} \le 3$ for all $n \in {\mathbb{Z}}^{+}$

Feb 7, 2017

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

Proof: ${t}_{n} - {t}_{n - 1} = \frac{3 n + 5}{n + 2} - \frac{3 \left(n - 1\right) + 5}{\left(n - 1\right) + 2}$

$= \frac{3 n + 5}{n + 2} - \frac{3 n + 2}{n + 1}$

$= \frac{\left(3 n + 5\right) \left(n + 1\right) - \left(3 n + 2\right) \left(n + 2\right)}{\left(n + 1\right) \left(n + 2\right)}$

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

$> 0$ for all $n \in {\mathbb{Z}}^{+}$

$\therefore {t}_{n} > {t}_{n - 1}$ for all $n \in {\mathbb{Z}}^{+}$

(b)

Proof: (by induction)

Base case: For $n = 1$, we have ${t}_{1} = \frac{8}{3} \in \left[\frac{8}{3} , 3\right]$.

Inductive hypothesis: Suppose that ${t}_{k} \in \left[\frac{8}{3} , 3\right]$ for some $k \in {\mathbb{Z}}^{+}$.

Induction step: We wish to show that ${t}_{k + 1} \in \left[\frac{8}{3} , 3\right]$. Indeed,

$\frac{8}{3} \le {t}_{k} \text{ }$ (by the inductive hypothesis)

$\le {t}_{k + 1} \text{ }$ (by the previous proof)

$= \frac{3 n + 8}{n + 3}$

$< \frac{3 n + 9}{n + 3}$

$= 3$

$\therefore {t}_{k + 1} \in \left[\frac{3}{8} , 3\right]$

We have supposed true for $k$ and shown true for $k + 1$, thus, by induction, ${t}_{n} \in \left[\frac{3}{8} , 3\right]$ for all $n \in {\mathbb{Z}}^{+}$. ∎