Let $x>0$.

For sequence $a_n$, such that $n$ denotes the $n$th term:

$$\begin{align} a_1&= \sqrt{x}\\ a_2&= \sqrt{x \sqrt{x}}\\ a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\ a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\ &\vdots\\ a_{n-1}&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x}}}}\\ a_n&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x \sqrt{x}}}}}\end{align}$$

How could one prove that:

$${a_n = x^{1-2^{-n}}}?$$

  • 4
    $\begingroup$ That's a sequence, not a series. $\endgroup$ – Hagen von Eitzen May 26 '13 at 21:58
  • 10
    $\begingroup$ $a_n=\sqrt{xa_{n-1}}$. Use induction. $\endgroup$ – Jonas Meyer May 26 '13 at 21:59

Hint: \begin{align} a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}} \end{align}

Can you use induction in these footsteps?


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