Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that $n<3^n$ where $n \in \mathbb N$,

when $ n=1 $, I have proved it's true.
And assumed when $n = p$ , $p<3^p$ is true.
Can any body help me in showing that it is true for $n =p+1$

share|cite|improve this question
Ya but the correct answer is not selected by the person who have asked the question and in my question it has no power in n. – Marlon Abeykoon Dec 7 '13 at 14:33
no power? $n=n^1$ – lab bhattacharjee Dec 7 '13 at 14:35

3 Answers 3

up vote 4 down vote accepted

We have as our inductive hypothesis (IH) that $\,p \lt 3^p$.

So we need to show $$p+1 \lt 3^{p+1} = 3 \cdot 3^p.$$

Can you see that $$p + 1 \quad \overset{IH}{\lt}\quad 3^p + 1 \quad \lt \quad 3\cdot 3^p \;= \;3^{p+1}, \quad \forall p \in \mathbb N$$

share|cite|improve this answer
Thanks, @Sami! ;-) – amWhy Dec 7 '13 at 14:31
yes i agree +1 also – dato datuashvili Dec 7 '13 at 14:35
Great! but in last line you have mentioned $(3^p)+1 < 3*(3^p)$. How can we say like that directly. – Marlon Abeykoon Dec 7 '13 at 14:44
@dato $-1\notin \mathbb N$! – amWhy Dec 7 '13 at 15:10
Not necessarily when we are working with inequalities: $3^p + \color{red}{1} \leq 3^p + \color{red}{3^p}$ because $1 \leq 3^p$, and for all $p > 0$, it happens to be true that $1 \lt 3^p$. This allows us then to say that $3^p + 1 \lt 3^p + 3^p \lt 3^p + 3^p + 3^p = 3\cdot 3^p = 3^{p+1}$. Note when $\lt$ is used, and note when $=$ is used. – amWhy Dec 7 '13 at 15:25

If $n < 3^n$ then it follows that $$n+1 <3^n +1 < 3^n +(2\cdot 3^n)=3^{n+1}$$ here I used that $1<2\cdot 3^n$ for all $n\in \Bbb N$.

share|cite|improve this answer

so we have proved that for $n=1$ ,$1<3$, what would be for $n+1$? we have $(n+1)<3^{n+1}$

which means that $(n+1)<3*3^n$ ,now clearly we have $3^p+1<3*3^p$, but also $3^p+1>p+1$ because we have used such thing that $p<3^p$, finally we conclude that $p+1<3*3^p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.