For what natural numbers n is the inequality $2^{n} > n^{2}$ valid?

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Solution

The inequality holds for $n = 1$ , Since $2^{1} > 1^{2}$
The inequality does not holds for $n = 2$ , Since $2^{2} = 2^{2}$
The inequality does not holds for $n = 3$ , Since $2^{3} < 3^{2}$
The inequality does not holds for $n = 4$ , Since $2^{4} = 4^{2}$
The inequality holds for $n = 5$ , Since $2^{5} > 5^{2}$
The inequality holds for $n = 6$ , Since $2^{6} > 6^{2}$
Now Assume,
$2^{m} > m^{2}$
Where m is a natural number $\geq 5$ .......................(1)
But problem 11, we have that
$2^{m} > 2 m + 1 f o r m \geq 3$ ............(2)
Adding the inequalities (1) and (2) we obtain,
$2^{m} + 2^{m} > m^{2} + + 2 m + 1$
that is, $2^{m + 1} > (m + 1)^{2}$
Hence, the inequality holds for $n = m + 1$
In follows by mathematical induction that the inequality
$2^{n} > n^{2}$ holds for $n = 1$ and for all natural numbers $n \geq 5$

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For what natural numbers n is the inequality 2n>n2 valid?

For what natural numbers n is the inequality $2^{n} > n^{2}$ valid?