3

Find all natural numbers $n$ such that $7n^3 < 5^n$.

I drew a graph which showed that $n \geq 4$

wolfram

How can I prove that? I guess I need to use induction with the base case $n=4$?

But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not substitute...

Belgi
  • 23,150
Justin D.
  • 745
  • 2
    $7n^3<5^n \implies 35n^3<5^{n+1}$. Now you have to show that $7(n+1)^3\le35n^3$ for suffienctly large $n$. I think you can take it from here. – chubakueno Sep 22 '13 at 21:14

2 Answers2

3

Base Case: For $n=4$, we have $7(4)^3 = 448 < 625 = 5^4$, which works.

Induction Hypothesis: Assume that the claim holds true for all $n \in \{4,\ldots,k\}$, where $k>3$.

It remains to prove the inequality true for $n=k+1$. Indeed, observe that: \begin{align*} 7(k+1)^3 &= 7(k^3 + 3k^2 + 3k + 1) \\ &< 7(k^3 + 3k^2 + 9k + 27) \\ &= 7(k^3 + (3)k^2 + (3)^2k + (3)^3) \\ &< 7(k^3 + (k)k^2 + (k)^2k + (k)^3) & \text{since }3 < k\\ &= 7(4k^3) \\ &= 4(7k^3) \\ &< 5(7k^3) \\ &< 5(5^k) & \text{by the induction hypothesis}\\ &= 5^{k+1}\\ \end{align*} as desired. This completes the induction.

Adriano
  • 41,576
1

First, check for $n = 4$:

$$7n^3 = 448 < 625 = 5^n.$$

Now, assume that it is true for all numbers strictly less than $n$. Let us check for $n$:

\begin{align*} 7n^3 &= 7((n-1)+1)^3 = 7(n-1)^3 + 7 \cdot 3 \cdot (n-1)^2 + 7 \cdot 3 \cdot (n-1) + 7. %\\ %&< 4 \cdot 7 (n-1)^3 < 4 \cdot 5^{n-1} < 5 \cdot 5^{n-1} = 5^n. \end{align*}

Here, use that $3(n-1)^k < (n-1)^3$ for $k \in \{1,2\}$, which comes from the fact that $n > 4$, so $n - 1 > 3$. Also, $7 < (n-1)^3$. If you get stuck, ask.

Vedran Šego
  • 11,372