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September 1st, 2019, 10:11 PM   #1
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Convergent or divergent

Show whether the expression converges .
$\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}}$.
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September 2nd, 2019, 10:28 AM   #2
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$\displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=1$.
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September 2nd, 2019, 02:14 PM   #3
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Quote:
Originally Posted by idontknow View Post
$\displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=1$.
$\displaystyle \displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} <n^{1/2}\cdot n^{1/4}\cdot n^{1/8} \cdot ... \cdot n^{1/2^n }<\underbrace{n^{1/2} \cdot n^{1/2} \cdot ...\cdot n^{1/2 }}_{n} =n^{1/2^n }$.
Since $\displaystyle \lim_{n\rightarrow \infty }n^{1/2^n } $ converges then $\displaystyle \displaystyle \displaystyle \displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}}$ converges .
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September 2nd, 2019, 02:51 PM   #4
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It appears to converge to about 1.6617
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September 2nd, 2019, 11:21 PM   #5
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Quote:
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It appears to converge to about 1.6617
How it comes to be greater than 1 while my method tells smaller than 1 ?(maybe inequalities i wrote are wrong)
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September 3rd, 2019, 12:12 AM   #6
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$\displaystyle \displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{\ln(n)}{2^n }}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{(\ln(n))'}{(2^n )\
' }}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{1}{n2^n\cdot \ln(2) }}=e^{0} =1$.
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September 3rd, 2019, 05:47 AM   #7
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Computer says no.

n = 1 to n = 100:
1.000000, 1.189207, 1.364262, 1.487738, 1.564477, 1.608895, 1.633541, 1.646864, 1.653947, 1.657670,
1.659612, 1.660620, 1.661140, 1.661407, 1.661544, 1.661615, 1.661651, 1.661669, 1.661678, 1.661683,
1.661685, 1.661687, 1.661687, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688,
1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688, 1.661688

I think the problem is you're forgetting it's an infinite product. The $\displaystyle n^{2^{-n}}$ terms go to one, but the product doesn't.
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