An arithmetic limit that gives the geometric mean

 Let $a_1, a_2, \dots a_k$ be positive reals.

Show that

$$ \lim_{n \to \infty} \left(\dfrac{a_1^{\frac{1}{n}} + \dots + a_k^{\frac{1}{n}}}{k}\right)^n = \sqrt[k]{a_1 a_2 \dots a_k}$$

Scroll down for a solution that does not use "calculus".

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A repeated use of (weighted) AM $\ge$ GM does it. 

Using weighted AM-GM we have that

$$\dfrac{x_1 \times \frac{1}{x_1} + \dots + x_k \times \frac{1}{x_k}}{x_1 + \dots + x_k} \ge \sqrt[x_1 + \dots + x_k] { \frac{1}{x_1^{x_1}} \dots \frac{1}{x_k^{x_k}}}$$

And so

$$ x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}} \ge \dfrac{x_1 + \dots + x_k}{k}$$

Where $S = x_1 + \dots + x_k$.

And so we have the following (by adding the classic AM-GM on the right) and taking the $n^{th}$ powers

$$\left(x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}}\right)^n \ge \left(\dfrac{x_1 + \dots + x_k}{k}\right)^n \ge \left(\sqrt[k]{x_1 \dots x_k}\right)^n$$

Now set $x_j = a_j^{\frac{1}{n}}$ (and so $x_j^n = a_j$)

Now as $n \to \infty, \frac{x_j}{S} \to \frac{1}{k}$.

Therefore 

$$\left(x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}}\right)^n = a_1^{\frac{x_1}{S}} \dots a_k^{\frac{x_k}{S}} \to \sqrt[k]{a_1 \dots a_k}$$

We also have

$$\left(\sqrt[k]{x_1 \dots  x_2}\right)^n =  \sqrt[k]{a_1 \dots a_k}$$

Thus $$ \left(\dfrac{a_1^{\frac{1}{n}} + \dots + a_k^{\frac{1}{n}}}{k}\right)^n = \left(\dfrac{x_1 + \dots + x_k}{k}\right)^n \to \sqrt[k]{a_1 \dots a_k}$$

by the sandwich theorem and we are done.