mersenneforum.org When does this series diverge?
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 2019-02-17, 05:57 #1 rudy235     Jun 2015 Vallejo, CA/. 312 Posts When does this series diverge? assume the series a ,aa, a^^aa 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 1.4222.05738816750076 ~ 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 e1/e Or the eth root of e =1.44466786100977 After about 190 iterations term a190 ~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 > e1/e WILL MAKE THE SERIES DIVERGENT?
 2019-02-17, 07:44 #2 rudy235     Jun 2015 Vallejo, CA/. 3C116 Posts I have proven heuristically that the highest number of a for which the series still converges is e1/e ~ 1.44466786100977 The term an n--> ∞ . is e (2.718281828...)
2019-02-17, 15:12   #3
Dr Sardonicus

Feb 2017
Nowhere

11·317 Posts

Quote:
 Originally Posted by rudy235 I have proven heuristically that the highest number of a for which the series still converges is e1/e ~ 1.44466786100977 The term an n--> ∞ . is e (2.718281828...)
Assuming the limit x satisfies

$a^{x} \;=\;x$

we have

$x^{\frac{1}{x}}\;=\;a\text{.}$

It is an easy exercise to prove that the largest value of

$y = x^{\frac{1}{x}}$

for positive real x (logarithmic differentiation works nicely) is

$y\;=\;e^{\frac{1}{e}}\text{.}$

This occurs at

$x\; =\; e\text{.}$

Last fiddled with by Dr Sardonicus on 2019-02-17 at 15:14 Reason: ginxif opsty

2019-02-17, 18:32   #4
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

52·173 Posts

Quote:
 Originally Posted by rudy235 assume the series a ,aa, a^^aa and succesively... For a=1 for instance, all terms are 1 thus the series is 1, 1, 1, .....1.1 it does not diverge.
If all the terms of a sequence are 1, the series obviously does diverge. 1 + 1 + 1 + 1+..... is not a finite sum.

Do you mean "sequence" everywhere that you said "series"?

2019-02-17, 20:57   #5
rudy235

Jun 2015
Vallejo, CA/.

312 Posts

Quote:
 Originally Posted by VBCurtis If all the terms of a sequence are 1, the series obviously does diverge. 1 + 1 + 1 + 1+..... is not a finite sum. Do you mean "sequence" everywhere that you said "series"?
Yep. I meant sequence. My bad!

2019-02-18, 16:13   #6
LaurV
Romulan Interpreter

Jun 2011
Thailand

8,753 Posts

Quote:
 Originally Posted by VBCurtis If all the terms of a sequence are 1, the series obviously does diverge. 1 + 1 + 1 + 1+..... is not a finite sum.
Hm... are you sure? :wondering:

I thought 1+1+1+1+....= -1/2

2019-02-18, 17:55   #7
Dr Sardonicus

Feb 2017
Nowhere

11·317 Posts

Quote:
 Originally Posted by LaurV Hm... are you sure? :wondering: I thought 1+1+1+1+....= -1/2
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 2-adic valuation! Under this "non-Archimedian" 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 p-adic 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 2019-02-18 at 18:02

 2019-02-18, 22:10 #8 wblipp     "William" May 2003 New Haven 2×32×131 Posts x = (1+1+1....) 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 2019-02-18 at 23:21 Reason: Correcting result
2019-02-18, 22:27   #9
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

9,127 Posts

Quote:
 Originally Posted by wblipp 1+x = 1+(1+1+1+1) = x
No, this only proves that 1 = 0, but tells us nothing about x. x cancels.

2019-02-18, 22:34   #10
chalsall
If I May

"Chris Halsall"
Sep 2002

246216 Posts

Quote:
 Originally Posted by Batalov No, this only proves that 1 = 0, but tells us nothing about x. x cancels.
For those of us who /try/ to understand what you guys talk about, could you please explain (in simple terms) how 1 == 0?

 2019-02-18, 23:26 #11 wblipp     "William" May 2003 New Haven 2×32×131 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 2019-02-18 at 23:35 Reason: add continuation video

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