20190217, 05:57  #1 
Jun 2015
Vallejo, CA/.
2×5×97 Posts 
When does this series diverge?
assume the series
a ,a^{a}, a^^a^{a} and succesively... For a=1 for instance, all terms are 1 thus the series is 1, 1, 1, .....1.1 it does not diverge. For a= SQRT(2) = 1.41421356.. the series is 1.41421356, 1.6325269, 76083956 ... and at approximately term 57 it converges to 2.00000000 This would be a great place to stop. However if a=1.42 it still seems to converge. At term 74 it seems to stop growing at 2.05738816750076 in other words 1.422^{2.05738816750076} ~ 2.05738816750076 I tried 1+4/9 =1.4444... This one takes longer but around term 135 it seems to stabilize at 2.63947300401328 My next try was e^{1/e} Or the eth root of e =1.44466786100977 After about 190 iterations term a_{190 } ~2.69004748029863 I believe this is the absolute limit for the series to converge. A slightly bigger number 1.445 diverges rather quickly. IS THERE SOME ANALYTICAL PROOF THAT ANY NUMBER > e^{1/e} WILL MAKE THE SERIES DIVERGENT? 
20190217, 07:44  #2 
Jun 2015
Vallejo, CA/.
2·5·97 Posts 
I have proven heuristically that the highest number of a for which the series still converges is e^{1/e} ~ 1.44466786100977
The term a_{n} n> ∞ . is e (2.718281828...) 
20190217, 15:12  #3  
Feb 2017
Nowhere
104F_{16} Posts 
Quote:
we have It is an easy exercise to prove that the largest value of for positive real x (logarithmic differentiation works nicely) is This occurs at Last fiddled with by Dr Sardonicus on 20190217 at 15:14 Reason: ginxif opsty 

20190217, 18:32  #4  
"Curtis"
Feb 2005
Riverside, CA
31×149 Posts 
Quote:
Do you mean "sequence" everywhere that you said "series"? 

20190217, 20:57  #5 
Jun 2015
Vallejo, CA/.
2×5×97 Posts 

20190218, 16:13  #6 
Romulan Interpreter
Jun 2011
Thailand
3^{4}×113 Posts 

20190218, 17:55  #7 
Feb 2017
Nowhere
5^{2}·167 Posts 
Hmm. I've seen the alternating series 1  1 + 1  1+ ... or Grandi's series being given the value 1/2, which answer can be obtained using Cesaro summation. But all terms +1, I don't know a reasonable way to assign a sum.
I do however, know a "standard" way to sum the geometric series 1 + 2 + 4 + 8 + ... and get 1, the sum given by blindly applying the usual formula for a geometric series 1 + x + x^2 + x^3 + ... = 1/(1  x). It's perfectly valid  if you use the 2adic valuation! Under this "nonArchimedian" valuation of the rationals, 2 = 1/2, so positive integer powers of 2 are "small," and the series is convergent. However, negative integer powers of 2 are now "large," so the geometric series 1 + 1/2 + 1/4 + ... is now divergent! EDIT: It suddenly occurs to me, under the padic valuation with p = 1/p, the series whose terms are all 1 is not only bounded, but has a subsequence of partial sums tending to 0. Every partial sum of a multiple of p terms is "small," of a multiple of p^2 terms even smaller, and so on. Alas, I'm too lazy to try and tease a sum out of this Last fiddled with by Dr Sardonicus on 20190218 at 18:02 
20190218, 22:10  #8 
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
1+x = 1+(1+1+1+1) = x 2x = 1 Ramanujan and others, I think. If I recall correctly, you can get to a reasonable interpretation through analytic continuation. oops. as Batalov points out, collecting the x's results in 0, not 2x. Last fiddled with by wblipp on 20190218 at 23:21 Reason: Correcting result 
20190218, 22:27  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·11·421 Posts 

20190218, 22:34  #10 
If I May
"Chris Halsall"
Sep 2002
Barbados
2·3·1,567 Posts 

20190218, 23:26  #11 
"William"
May 2003
New Haven
938_{16} Posts 
I was right about analytic continuation, but not the cheat for 1+1+1 ... = 1/2
Wikipedia has reasonable starting point And this is a good introduction to Analytic Continuation, especially of the Rieman Zeta function Last fiddled with by wblipp on 20190218 at 23:35 Reason: add continuation video 
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