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Problem: $2^n > n^2, \forall n \in \mathbb{N} , n \geq 5$

Base: $2^5 > 5^2$

Induction Hypothesis: Assume for $n = k \geq 5$ that $2^k>k^2$

Inductive Step: $$2^k > k^2$$ $$2^k \times 2 > k^2 \times 2$$ $$2^{k+1} > 2k^2$$

From there I can finish the proof by asserting that $k^2 > 2k+1, \forall k \in \mathbb{N} , k \geq 3$.

Do I need to prove that $k^2 > 2k+1, \forall k \in \mathbb{N} , k \geq 3$. Or can I substitute it into my inequality?

Arctic Char
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HopelessSoul
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4 Answers4

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You can use $$2^{k+1}=2^k+2^k> k^2 +2k+1$$

zkutch
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You need to prove $k^2\color{blue}{\ge}2k+1$ for $k\ge\color{blue}{5}$ (unless your teacher thinks that's obvious enough to skip), although the stronger version you quoted is also correct. The proof does not, however, need induction. because it's equivalent to $(k-1)^2\ge2$.

J.G.
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  • Would you be able to explain how it can be simplified to $(k-1)^2 \geq 2$? I understand how you could simplify it to $(k-1)^2 \geq 0$ but not how it becomes 2 – HopelessSoul Jul 17 '20 at 10:27
  • @HopelessSoul $k^2\ge2k+1$ rearranges to $(k-1)^2\ge2$ which, while not quite as trivial as $(k-1)^2\ge0$, follows if $k\ge3$ since then $(k-1)^2\ge2^2=4>2$. – J.G. Jul 17 '20 at 10:32
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You do need to prove it, as the onus is on you to demonstrate that you have understood the question.

My favourite method to prove this is to consider $2+\frac1k\le3$ for all $k\ge1$, and multiply through by $k$, then, as $k\ge3$, the RHS becomes $3k\le k^2$.

JMP
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Still another way:

If $n\ge 5$, we have $$\Bigl(\dfrac{n+1}n\Bigr)^{\!2}=\Bigl(1+\dfrac1n\Bigr)^{\!2}<\Bigl(1+\dfrac15\Bigr)^{\!2}<2,$$ whence, by the inductive hypothesis, $$2^{n+1}=2^n\cdot 2>n^2\cdot 2>n^2\Bigl(\dfrac{n+1}n\Bigr)^{\!2}=(n+1)^2.$$

Bernard
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