Mathematical induction problem $\frac{3^{n+1}}{3^n + 1} \gt 1$

Question

So in solving an exercise on mathematical induction i got stuck when trying to prove this.

$\forall n \in \mathbb{Z} $ with $ n \ge 0: $ $$\frac{3^{n+1}}{3^n + 1} \ge 1$$

The original problem is this:

$\forall n \in \mathbb{Z} $ with $ n \ge 0: $ $$ n < 3^n $$ I started solving by adding 1 on each side. $$ n + 1 < 3^n + 1 $$ multiply large side of inequality by something bigger than or equal to 1 $$ n + 1 < (3^{n}+1) \frac{3^{n+1}}{3^n +1}$$ $$ n + 1 < 3^{n+1}$$ Maybe my approach is wrong? Please help me out.

Answer 1

0) $3^n \gt n$ for $n=0,1$, ok.

Let $n \in \mathbb{Z^+},$ $ n \ge 1.$

1) Hypothesis $: 3^n \gt n $.

2) Step: $n+1;$

Multiply both sides by $3:$

$3^{n+1} \gt 3n = n +2n \gt n +1$ for $n\ge 1$.

Answer 2

Basically, we need to prove: $$P_n: n<3^n \tag 1$$ Note that $P_1$ holds. So, we need to prove the validity of $P_{n+1}$. For that note: $$n < 3^n$$ $$\implies n+1 <3^n +1 < 3^n +3^n +3^n = 3^n\times 3 = 3^{n+1}$$

Hence, as $P_{n+1}$ holds, we can say that $P_n$ holds for all $n \geq 0$.

Answer 3

For $n = 0$, we have $\frac{3}{2} \ge 1$ so argument holds. Then assume, inductively, that $n \ge 1$ and argument holds for all $n$. Then, for $n+1$, we have $$\frac{3^{n+2}}{3^{n+1}+1} = \frac{3 \cdot 3^{n+1}}{3 \cdot 3^n+1} \ge \frac{3 \cdot 3^{n+1}}{3 \cdot 3^n+3} = \frac{3^{n+1}}{3^n+1}$$ and by inductive assumption, we already have $\frac{3^{n+1}}{3^n+1} \ge 1$ so $\frac{3^{n+2}}{3^{n+1}+1} \ge 1$ which means argument holds for $n+1$. Therefore, by induction, argument holds for all $n \ge 0$.

Answer 4

Alternative way, so we have $$\frac{3^{n+1}}{3^n+1} \iff 3^{n+1} > 3^n+1 \iff 3^{n}+3^{n}+3^{n} > 3^n+1 \iff \color{red}{2\cdot3^{n}>1}$$

The red part is easy to prove by induction:

• $n=0 \Rightarrow 2>1$
• Assuming it is true for $n$ then $2\cdot 3^n>1 \Rightarrow 2\cdot 3^{n+1}>3>1 \Rightarrow 2\cdot 3^{n+1} > 1$