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How do I prove the sequence $\{\sqrt{7}, \sqrt{7\sqrt{7}},\sqrt{7\sqrt{7\sqrt{7}}}{,... \}}$ converges at 7? I understand intuitively that the final term would be $7^{1/2} \cdot7^{1/4} \cdot7^{1/8}\ldots$ , and that would converge ultimately to $7^1$ but I'm not sure how to properly show that. Thanks!

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    $\begingroup$ The $n$-th term of the sequence is $7^{\sum_{k=1}^n2^{-k}}.$ Now use the fact that $\sum_{k=1}^n2^{-k}$ is a geometric series to conclude that your sequence converges to $7.$ $\endgroup$ – CIJ Dec 1 '16 at 21:57
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use the geometric series $${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$ $$\{\sqrt{7\sqrt{7\sqrt{7}}}{... \}}=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}=7^{\frac{1}{2}(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...)}=7^{\frac{1}{2}\frac{1}{1-\frac{1}{2}}}=7^1$$

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$$a_1 = \sqrt{7}$$ $$0<a_1 < 7$$ $$a_{n+1} = \sqrt{7a_{n}}$$ By geometric mean, $$a_{n} < a_{n+1} < 7$$

Hence, $a_{n}$ is monotonic increasing and bounded above

$$\sqrt{7} \le a_{n} < a_{n+1} < 7$$ $$7-a_{n+1}=\frac{\sqrt{7}(7-a_{n})}{\sqrt{7}+\sqrt{a_{n}}}$$ $$0<7-a_{n+1} \le \frac{\sqrt{7}(7-a_{n})}{\sqrt{7}+\sqrt{a_{1}}}$$ $$0<7-a_{n+1} \le \left( \frac{\sqrt{7}}{\sqrt{7}+\sqrt[4]{7}} \right)^n (7-a_{1})$$ Since $0<\dfrac{\sqrt{7}}{\sqrt{7}+\sqrt[4]{7}}<1$,

$$7-a_{n+1} \to 0 \quad \text{as} \quad n\to \infty$$

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