For the sake of a quick, "pure" math post.
It is a well known fact that \(\lim_{n \to \infty} \sqrt[n]{n} = 1\)
There are multiple proofs of that. One such proof is to use the inequality of Arithmetic and Geometric means
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1\cdot x_2\cdots x_n} \quad (x_i \ge 0)$$
Choose \(x_1 = x_2 = \dots = x_{n-2} = 1\) and \(x_{n-1} = x_n = \sqrt{n}\)
Then we get,
$$\frac{n-2 + 2\sqrt{n}}{n} \ge \sqrt[n]{n} \ge 1 $$
And so,
$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge \sqrt[n]{n} \ge 1$$
By the sandwich theorem, the limit is one.
It is a well known fact that \(\lim_{n \to \infty} \sqrt[n]{n} = 1\)
There are multiple proofs of that. One such proof is to use the inequality of Arithmetic and Geometric means
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1\cdot x_2\cdots x_n} \quad (x_i \ge 0)$$
Choose \(x_1 = x_2 = \dots = x_{n-2} = 1\) and \(x_{n-1} = x_n = \sqrt{n}\)
Then we get,
$$\frac{n-2 + 2\sqrt{n}}{n} \ge \sqrt[n]{n} \ge 1 $$
And so,
$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge \sqrt[n]{n} \ge 1$$
By the sandwich theorem, the limit is one.
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