$(1+\frac{1}{3})^n \ge 1+\frac{n}{3}$, $\forall n\geq 0$ proof by induction

Question

How can I prove this by induction?
Here's the question

Step 1 ) for $n = 0$ both it's true
Step 2 ) for $n= k$ assume that $\textstyle\displaystyle{(1+\frac{1}3)^k\geq 1 + \frac{k}{3}}$ is true
Step 3 ) prove that for $n = k+1$, $\textstyle\displaystyle{(1+\frac{1}3)^k+1$ \geq 1 + \frac{k+1}{3}}$ is true
I started with the left side $(1+\frac{1}3)^{k+1}$ = $(1+\frac{1}3)^k(1+\frac{1}3)$
then I replaced $\textstyle\displaystyle{(1+\frac{1}{3})^k}$ with my assumption $\textstyle\displaystyle{1 + \frac{k}{3}}$ so now it's $\textstyle\displaystyle{(1+\frac{1}3)^k(1+\frac{1}3)\geq (k+\frac{1}3)(1+\frac{1}3)}$

What should I do next ?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Oct 06 '21 at 06:04
You're pretty much done if you can show $\left(1 + \frac{k}{3}\right)\left(1 + \frac{1}{3}\right) \ge 1 + \frac{k + 1}{3}$. Try expanding the left hand side, and see if any terms cancel on the right. — Theo Bendit, Oct 06 '21 at 06:13
Does this answer your question? Proof by induction of Bernoulli's inequality: $(1 + x)^n \geq 1 + nx$. There are quite a few other similar induction proof questions here, e.g., Bernoullis inequality proof, Bernoulli's inequality by induction and Prove Bernoulli's inequality. — John Omielan, Oct 06 '21 at 06:20
@Seba I mean just expand the brackets on the left hand side. I could tell you what it is, but I'd rather guide you to do it for yourself. What do you get when you expand $\left(1 + \frac{k}{3}\right)\left(1 + \frac{1}{3}\right)$? — Theo Bendit, Oct 06 '21 at 06:24
@Seba Can't argue with that! I was thinking more as presenting it unsimplified as a sum of four terms, i.e. $1 + \frac{1}{3} + \frac{k}{3} + \frac{k}{9}$. Compare to the right hand side: $1 + \frac{k + 1}{3}$. What do you notice on that's on the left hand side that isn't on the right hand side? Is there anything on the right hand side that isn't on the left hand side? Another way of thinking about it: take the difference $1 + \frac{1}{3} + \frac{4k}{9} - \left(1 + \frac{k+1}{3}\right)$. What is the difference, once fully simplified? Is it $\ge 0$? — Theo Bendit, Oct 06 '21 at 06:32
@TheoBendit if I take the difference it will be 1/3 +(k-3) / 9 and in the assumption k >=0 so in this way, I'm done ?? — Seba, Oct 06 '21 at 06:39
@TheoBendit or if I want to compare the right hand side is now 1+(k+1)/3 +k/9 so it will be definitely bigger than 1+(k+1)/3 which is the right hand side is this what you mean ?? — Seba, Oct 06 '21 at 06:43
@Seba Yes, either line of reasoning will work. Either which way, you've proven the given inequality. — Theo Bendit, Oct 06 '21 at 06:45
@TheoBendit thanks for everything it's actually the first time I feel like someone is trying to explain it to me step by step. — Seba, Oct 06 '21 at 06:48
@Seba No worries. To be fair, step-by-step is tricky in Q&A format. Glad I could help. — Theo Bendit, Oct 06 '21 at 06:56

score 1 · Accepted Answer · answered Oct 06 '21 at 07:32

Set $P(n): (1+\frac 13)^n\geq 1+\frac n3, n\in \Bbb N$. We see that $P(1): \frac 43\geq \frac 43$ is true. Suppose that $P(n)$ holds for some $n\in \Bbb N$. Then $(1+\frac 13)^{n+1}=(1+\frac 13)^n(1+\frac 13)\geq (1+\frac n3)(1+\frac13)=1+\frac13 +\frac n3 +\frac n6=1+(\frac 13 +\frac n3)+\frac n6\geq 1+\frac 13 (n+1): P(n+1).$

q.e.d.

$(1+\frac{1}{3})^n \ge 1+\frac{n}{3}$, $\forall n\geq 0$ proof by induction

1 Answers1