March 26th, 2017, 03:59 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  Sequence of "stacked" square roots
Hello all. I am working on a sequence problem involving square roots. Here is my work: http://i.imgur.com/N3If5GR.jpg I need to figure out if it converges or diverges and then the value if it converges. I have been trying to figure out a formula for the denominator of the exponent so it matches the respected root in each term, but I can't figure it out. Is there another way to do this without trying to find a formula for the nth term? Thanks! 
March 26th, 2017, 04:36 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,255 Thanks: 681 
It seems pretty clear the sequence converges to 1. It's like repeatedly punching the square root button on your calculator. What is the 16th root of 2? It's 1.000something. It has to be bigger than 1 but not much bigger.

March 26th, 2017, 04:43 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Look at the logarithm (base 2) of each element if Maschke's statement is either not obvious enough or not rigorous enough.

March 26th, 2017, 05:40 PM  #4 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  
March 26th, 2017, 06:43 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle 2^{1/3} \to log_2(~2^{1/3}) = 1/3$ Do that to each term. Does this give you any ideas? Dan  
March 26th, 2017, 06:44 PM  #6 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics 
Got it now. Formula for the nth term is 2^[1/2^(n1)] Taking the limit as n approaches infinity equals 1. 

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roots, sequence, square, stacked 
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