Prove by induction the particular inequality $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$

Question

$\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$

Not sure where I'm going wrong in my Algebra, but I assume it's because I'm adding an extra term. Is the extra term unnecessary because it's not a summation I'm trying to prove?

Starting the proof:

$n=1 \Rightarrow \left(1.3\right)^{1} \ge 1 + \left(0.3\right)(1) \Rightarrow 1.3 \ge 1.3$

$n=k \Rightarrow \left(1.3\right)^{k} \ge 1 + \left(0.3\right)(k) \Rightarrow 1.3 \ge 1.3$

$n=k+1 \Rightarrow \left(1.3\right)^{k}+\left(1.3\right)^{k+1} \ge 1 + \left(0.3\right)(k+1) \Rightarrow 1.3 \ge 1.3$

After trying some Algebra I end up with this:

$= 1 + \left(0.3\right)k+\left(1.3\right)\left(1.3\right)^{k} \ge 1 + \left(0.3\right)k + .3$

Any hints?

Answer 1

I'm not sure what you mean by “$\implies 1.3\ge1.3$”.

---

The proof should go like this.

1. The statement is true for $k=1$; indeed, $(1.3)^1=1.3=1+0.3\cdot1$

2. Suppose the statement holds for $n=k$, that is, $(1.3)^k\ge1+0.3k$; then $$ (1.3)^{k+1}=(1.3)^k\cdot 1.3\ge(1+0.3k)\cdot(1+0.3)=1+0.3k+0.3+0.9k\ge 1+0.3(k+1) $$ so the statement holds also for $n=k+1$.

Actually, it's even easier to prove that, for $r>0$, $$ (1+r)^n\ge1+rn $$ and yours is the special case for $r=0.3$. (The hypothesis $r>0$ can actually be relaxed to $r>-1$.) This is known as Bernoulli inequality.

Answer 2

(EDIT: I realized after I posted that you asked for a proof by induction. My bad! The earlier answer posted by egreg does this, and is much simpler in my opinion.)

We want to show that $1.3^n \geq 1 + 0.3n$ for all $n > 0$. Let's look only at the left side for now. We can write $$ 1.3^n = (1 + 0.3)^n.$$ Then by the binomial theorem we have that $$ 1.3^n = (1 + 0.3)^n = \sum_{k=0}^n \binom{n}{k} 0.3^k = 1 + 0.3 n + \sum_{k=2}^n \binom{n}{k} 0.3^k,$$ where we pulled the first two terms out of the summation so that we can directly compare it to the right side of the inequality. Thus, we have that $1.3^n \geq 1 + 0.3n$ provided $\sum_{k=2}^n \binom{n}{k} 0.3^k \geq 0$, which we know is true since all of the involved quantities are positive (or zero).

Answer 3

This is true in more general form: $$ (1+x)^n \geq 1+ nx,\quad x\geq 0$$

The proof is just the binomial theorem:

$$(1+x)^n = 1 + nx + \frac{n(n-1)}2x^2+\cdots \geq\, 1 + nx$$ just let $x = 0.3$.

This is obvious for $x\geq 0$, and for $x\geq -1$ is known as Bernoulli's inequality, as noted in the answer by egreg.

Answer 4

Let's prove Bernoulli's inequality by induction. Assume $x\ge -1$.

Base step $n=1$:

$$(1+x)^1 = 1+1\cdot x.$$

Now induction step. Assume $(1+x)^n \ge 1+nx$. Then

\begin{align*} (1+x)^{n+1} &= (1+x)^n (1+x) \\ & \ge (1+nx) (1+x)\\ & = 1+(n+1)x +nx^2\\ &\ge 1+(n+1)x. \end{align*}

Note that the first inequality holds because of the induction assumption and since $x\ge -1$.

Answer 5

I want to comment on the use of $\Rightarrow$. Usually it is used to denote implications, not equivalences. Also the arrow should point into the direction of which statement you want to prove. Further $\Rightarrow$ is rather surrounded by statements, than by equations. For example "$X$ solves the equation x^2=3" and not only $x^2=3$. It is because $\Rightarrow$ relates statements which are true or false and an equation without context is not always true or false it depends on the context.

You wrote $n=k \Rightarrow \cdots \Rightarrow 1.3 \ge 1.3$ those "statements" are not equivalent.

With the equations $n=1$, $n=k$ and $n=k+1$ you want to label the different cases. So it is a list of cases and I think it would be better to write $n=1:$, $n=k:$, $\dots$. For $n=1$ this would look like $$ n=1: \quad (1.3)^1 = 1.3 \ge 1.3 = 1 + (0.3)1. $$

To get from $n=k$ to $n=k+1$ you should take the equation for $n=k$ and try to get the equation for $n=k+1$ out of it. In this case multiplication is a good choice: $$ n=k+1: \quad (1.3)^{k+1} = 1.3 (1.3)^k \ge 1.3 (1+ (0.3)k)= 1+ 0.3 + 1.3(0.3)k \ge 1 +(0.3)(k+1). $$

Answer 6

Just an other way

If you are interested by an other proof, $$(1.3)^n-1=(1.3)^n-1^n=(1.3-1)(1+1.3+(1.3)^2+...+(1.3)^{n-1})=0.3(1+\underbrace{1.3}_{\geq 1}+\underbrace{(1.3)^2}_{\geq 1}+...+\underbrace{(1.3)^{n-1}}_{\geq 1})\geq 0.3(\underbrace{1+...+1}_{n\ times })=0.3n$$ what prove the claim.