Prove using induction : $n < 3^n$

Question

$ p(n) = n < 3^n = q(n) $

when $n=1$, $p(n)=1< 3=q(n)$

Assume the result is true for $n=m$

$p(m)=m < 3^m$

when $n = m+1$
$p(m+1) = m+1 < 3^m +1<3*3^m = 3^{m+1}=q(m+1)$

is this correct?

I tried up to this extent to solve this. And also tried other methods. But no luck. Can some one show me the way to get this done.

Thank you.

And what can you say about $3^p+1$ and $3^{(p+1)}$? Consider that $3^{(p+1)}=3^p 3=3^p+3^p+3^p$ and for $p>1$ is trivial to see that $3^p 2>1$. This should be enough to help you... — Umberto, Dec 23 '13 at 16:52
If it helps, you can find several similar questions on this site; for example: http://math.stackexchange.com/questions/449672/prove-that-n-2n-for-all-natural-numbers-n, http://math.stackexchange.com/questions/169859/prove-by-mathematical-induction-that-2n-2n-for-all-integer-n1 In fact, this one is about exactly the same inequality: http://math.stackexchange.com/questions/596876/prove-inequality-3nn-using-mathematical-induction — Martin Sleziak, Dec 23 '13 at 20:57

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

Yes, your proof correct. There is a more general way to proceed that lends further insight.

Hint $\ $ We seek an inductive proof of $\:\dfrac{3^n}n > 1.\,$ True for $\, n=1.\,$ For the inductive step, note

$$\quad \dfrac{3^{n+1}}{n+1}\, =\, \color{#0a0}{\dfrac{3\cdot n}{n+1}} \color{#c00}{\dfrac{3^n}n} > 1$$

since red $\rm\color{#c00}{term} > 1$ by induction, $ $ and the green $\,\rm\color{#0a0}{term} > 1\,$ by $\, n>1 \Rightarrow 3n > n\! +\! 1.\ \ $ QED

Remark $\, $ Unwinding the induction yields a proof by multiplicative telescopy, e.g.

$$ \dfrac{3^5}5 =\, \dfrac{3\cdot\color{#c00}4}{5\quad\ } \dfrac{3\cdot\color{#0a0}3}{\color{#c00}4\quad\ } \dfrac{3\cdot\color{orange}2}{\color{#0a0}3\quad\ }\dfrac{3\cdot\color{brown}1}{\color{orange}2\quad\ }\dfrac{3}{\color{brown}1} $$

Note how same-color terms "telescopically" cancel, leaving five $3$'s on top, and $5$ on the bottom. Since each fraction is $> 1$ so too is their product (a trivial induction). In essence, telescopy simplifies the induction to a trivial induction, that a product of terms $> 1$ is $> 1$.

Similarly, telescopic methods often serve to greatly simply inductive proofs involving not only powers but also products, e.g. factorials. Similarly, there is also additive telescopy for sums.

1

Excellent remark on telescopy. – Newb Dec 23 '13 at 20:53

score 4 · Answer 2 · answered Dec 23 '13 at 16:51

4

For $n=1$ the statement is clear. Hence we assume that $n\leq 3^n$ holds for $n$ and we want to show the formula for $n+1$.

$$n+1 \leq 3^n +1 \leq 3^n +3 \leq 3^{n+1}$$ Do you see why each estimate is true?

answered Dec 23 '13 at 16:51

Quickbeam2k1

2,139

1

$$3^n+1 < 3^n +3 $$ – Adi Dani Dec 23 '13 at 21:17
sorry for proving a weaker result :) – Quickbeam2k1 Dec 23 '13 at 22:48

score 2 · Answer 3 · answered Dec 23 '13 at 20:42

2

The correct answer is YES. Your answer was correct and it seems people didn't actually read your post because it was messy. I've cleaned it up and it was clearly correct

answered Dec 23 '13 at 20:42

Squirtle

6,698

I can see Martin Sleziak edited it since... – Squirtle Dec 23 '13 at 21:14

score 0 · Answer 4 · answered Dec 23 '13 at 16:52

For $n=1$ it is true and $1<3^1=3$.

Assume now that for a $k\ge 1$, we have that $k<3^k$. Then $3k<3\cdot 3^k=3^{k+1}$. But $3k>2k+1>k+1$, and altogether $$ k+1<3k<3^{k+1}. $$ Thus the inductive hypothesis $\forall k\big(p(k)\Rightarrow P(k+1)\big)$ is valid, and so is the inequality.

Prove using induction : $n < 3^n$

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