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-   -   When does this series diverge? (https://www.mersenneforum.org/showthread.php?t=24093)

rudy235 2019-02-17 05:57

When does this series diverge?
 
assume the series
a ,a[SUP]a[/SUP], a^^a[SUP]a[/SUP] 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[SUP]2.05738816750076[/sup] ~ 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[SUP]1/e[/SUP] Or the eth root of e =1.44466786100977

After about 190 iterations term a[SUB]190 [/SUB] ~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[SUP]1/e[/SUP] WILL MAKE THE SERIES DIVERGENT?

rudy235 2019-02-17 07:44

I have proven heuristically that the highest number of a for which the series still converges is e[SUP]1/e[/SUP] ~ 1.44466786100977
The term a[SUB]n[/SUB] n--> ∞ . is e (2.718281828...)

Dr Sardonicus 2019-02-17 15:12

[QUOTE=rudy235;508780]I have proven heuristically that the highest number of a for which the series still converges is e[SUP]1/e[/SUP] ~ 1.44466786100977
The term a[SUB]n[/SUB] n--> ∞ . is e (2.718281828...)[/QUOTE]
Assuming the limit x satisfies

[tex]a^{x} \;=\;x[/tex]

we have

[tex]x^{\frac{1}{x}}\;=\;a\text{.}[/tex]

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

[tex]y = x^{\frac{1}{x}}[/tex]

for positive real x (logarithmic differentiation works nicely) is

[tex]y\;=\;e^{\frac{1}{e}}\text{.}[/tex]

This occurs at

[tex]x\; =\; e\text{.}[/tex]

VBCurtis 2019-02-17 18:32

[QUOTE=rudy235;508777]assume the series
a ,a[SUP]a[/SUP], a^^a[SUP]a[/SUP] and succesively...

For a=1 for instance, all terms are 1
thus the series is 1, 1, 1, .....1.1 it does not diverge.
[/QUOTE]

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"?

rudy235 2019-02-17 20:57

[QUOTE=VBCurtis;508812]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"?[/QUOTE]
Yep. I meant sequence. My bad!

LaurV 2019-02-18 16:13

[QUOTE=VBCurtis;508812]If all the terms of a sequence are 1, the series obviously does diverge. 1 + 1 + 1 + 1+..... is not a finite sum.
[/QUOTE]
Hm... are you sure? :wondering:

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

Dr Sardonicus 2019-02-18 17:55

[QUOTE=LaurV;508867]Hm... are you sure? :wondering:

I thought 1+1+1+1+....= -1/2
:razz:[/QUOTE]

Hmm. I've seen the alternating series 1 - 1 + 1 - 1+ ... or [url=https://en.wikipedia.org/wiki/Grandi%27s_series]Grandi's series[/url] being given the value 1/2, which answer can be obtained using [url=https://www.quora.com/What-is-the-significance-of-Cesaro-summation]Cesaro summation[/url]. 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 -- [i]if[/i] 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, [i]negative[/i] integer powers of 2 are now "large," so the geometric series

1 + 1/2 + 1/4 + ...

is now divergent!

[b]EDIT:[/b] 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
:yawn:

wblipp 2019-02-18 22:10

[strike]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.[/strike]

oops. as Batalov points out, collecting the x's results in 0, not 2x.

Batalov 2019-02-18 22:27

[QUOTE=wblipp;508902]
1+x = 1+(1+1+1+1) = x
[/QUOTE]
No, this only proves that 1 = 0, but tells us nothing about x. x cancels.

chalsall 2019-02-18 22:34

[QUOTE=Batalov;508903]No, this only proves that 1 = 0, but tells us nothing about x. x cancels.[/QUOTE]

For those of us who /try/ to understand what you guys talk about, could you please explain (in simple terms) how 1 == 0? :smile:

wblipp 2019-02-18 23:26

I was right about analytic continuation, but not the cheat for 1+1+1 ... = -1/2

[URL="https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF"]Wikipedia has reasonable starting point[/URL]

And [URL="https://www.youtube.com/watch?v=sD0NjbwqlYw"]this[/URL] is a good introduction to Analytic Continuation, especially of the Rieman Zeta function


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