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 September 1st, 2019, 10:11 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Convergent or divergent Show whether the expression converges . $\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}}$.
 September 2nd, 2019, 10:28 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 $\displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=1$.
September 2nd, 2019, 02:14 PM   #3
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Quote:
 Originally Posted by idontknow $\displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=1$.
$\displaystyle \displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} <n^{1/2}\cdot n^{1/4}\cdot n^{1/8} \cdot ... \cdot n^{1/2^n }<\underbrace{n^{1/2} \cdot n^{1/2} \cdot ...\cdot n^{1/2 }}_{n} =n^{1/2^n }$.
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 .

 September 2nd, 2019, 02:51 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 It appears to converge to about 1.6617 Thanks from idontknow
September 2nd, 2019, 11:21 PM   #5
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Quote:
 Originally Posted by romsek 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)

 September 3rd, 2019, 12:12 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 $\displaystyle \displaystyle 0 <\displaystyle \sqrt{1\sqrt{2 \sqrt{3\cdot ....\cdot ...}}} < \lim_{n\rightarrow \infty } \sqrt[\displaystyle 2^n ]{n}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{\ln(n)}{2^n }}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{(\ln(n))'}{(2^n )\ ' }}=e^{\lim_{n\rightarrow \infty}\displaystyle \frac{1}{n2^n\cdot \ln(2) }}=e^{0} =1$.
 September 3rd, 2019, 05:47 AM #7 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 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. Thanks from idontknow

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