About an alternating sign series...

**chisigma** · May 9th 2009, 12:46 AM

Let's consider the alternating sign series...

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ (1)

If we indicate with $a_{n}$ the general term of (1), you can verify that...

1) $\{|a_{n}|\}$ is a decreasing sequence

2) $\lim_{n \rightarrow \infty} a_{n}=0$

... so that (1) converges. Now, indicating with $\sigma$ the sum of (1), the questions I intend to indicate to your attention are...

a) is $\sigma$ rational or irrational?...

b) if $\sigma$ is irrational, can it be expressed as elementary function of rational numbers or other constants like $\pi$ , $e$ and so one?...

Kind regards

$\chi$ $\sigma$

**Media_Man** · May 11th 2009, 10:25 AM

The Dirichlet Eta function: $\eta (s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$

Therefore, your query, $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}=\eta(\frac{1}{2})$

It is known that $\eta(s)=(1-2^{1-s})\zeta(s)$ , so $\eta(\frac{1}{2})=(1-\sqrt{2})\zeta(\frac{1}{2})=.60489864341767...$

"It is apparently not known if the value $\zeta(\frac{1}{2})=-1.46035450880...$ (Sloane's A059750) can be expressed in terms of known mathematical constants." Riemann Zeta Function -- from Wolfram MathWorld

Are you attempting to resolve this currently open question?

**chisigma** · May 12th 2009, 04:46 AM

May be that what Media_Man has 'suspected' is not far from the true

... in any case we have nothing to lose in attempting

... in order to have some raisonable chanche let's spend a little time explaining something about the so called 'Euler transform'...

Given a sequence $a_{n}$ , we define these two operators...

$\Delta a_{n} = a_{n+1}- a_{n}$

$E a_{n} = a_{n+1}$ (1)

The fundamental relation between $\Delta$ and $E$ is easy to find. Considering that...

$(1 + \Delta) a_{n} = a_{n}+a_{n+1}-a_{n} = a_{n+1} = E a_{n}$

... you fast arrive to conclude that is...

$\Delta = E -1 \rightarrow E= 1 + \Delta$ (2)

From (2) we derive for instance that...

$\Delta^{2} a_{n} = \Delta(\Delta a_{n})= \Delta (a_{n+1}- a_{n})= a_{n+2} -2\cdot a_{n+1} + a_{n} = E^{2}a_{n} -2\cdot E a_{n} + a_{n} \rightarrow$

$\rightarrow \Delta^{2}= E^{2} -2 E + 1$

... and, using (1), (2) and the binomial series expansion, we arrive to the following general relations...

$\Delta^{n} = \sum_{k=0}^{n} (-1)^{k}\binom{n}{k}E^{n-k}$

$E^{n} = \sum_{k=0}^{n} \binom{n}{k}\Delta^{k}$ (3)

Let's consider now the alternating sign series...

$S= a_{0} - a_{1} + a_{2} - a_{3} + \dots = \sum_{n=0}^{\infty}(-1)^{n} a_{n}$

On the basis of the results we have now obtained in the XVIII° century the Swiss mathematician Leonhard Euler supposed that this identity was true...

$S= \sum_{n=0}^{\infty}(-1)^{n} a_{n}= \frac{a_{0}}{1+E}= \frac{a_{0}}{2+\Delta}= \frac{1}{2}\cdot \frac{a_{0}}{1+\frac{\Delta}{2}}=$

$= \frac{1}{2}\cdot (a_{0} - \frac{\Delta a_{0}}{2} + \frac{\Delta^{2}a_{0}}{4} - \frac{\Delta^{3}a_{0}}{8} +\dots)=\frac{1}{2} \sum_{n=0}^{\infty}(-1)^{n}\cdot \frac{\Delta^{n}a_{0}}{2^{n}}=$

$=\frac{1}{2} \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2^{n}} \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k}\cdot a_{n-k}$ (4)

Today it doesn't jet exist a general criterion to establish when (4) is valid and when it isn't... but why not to try?...

Kind regards

$\chi$ $\sigma$

**chisigma** · May 12th 2009, 11:49 PM

Yesterday we are arrived to describe the 'accelerated procedure' that Euler 'discovered' two and half centuries ago for the computation of alternating sign series...

$\sum_{n=0}^{\infty} (-1)^{n} a_{n}= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} a_{n-k}$ (1)

Today we will perform the 'next logical step': the attempt to use (1) to compute the sum of the alternating sign series...

$\sigma = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\sqrt{n+1}}$ (2)

It is relatively easy to see that a 'brute force attack' to (2) requires about $10^4$ iterations for a two digits precision and 'only' about $2\cdot 10^{6}$ iterations [!!...] for a three digits precision

...

The situation becomes much more favorable if we use the 'Euler's procedure'. If we set $a_{n}= \frac{1}{\sqrt{n+1}}$ in the second term of (1) we obtain two digits precision in 3 iterations , three digits precision in 5 iterations and twelve digits precision in 35 iterations and is...

$\sigma= .604898643422...$

I suppose it is not exaggerated if I say that the 'Euler's solution' , at least in this case, does work exellently!

... the questions I posed about $\sigma$ are however still open... but... never say never again!!...

Kind regards

$\chi$ $\sigma$

**Media_Man** · May 13th 2009, 06:00 AM

I did not mean to discourage your hunt, only to bring the issue up to speed on current progress. The closest known result to your question of the value of $\sigma$ that I have found is this:

$\frac{\zeta'(\frac{1}{2})}{\zeta(\frac{1}{2})}=\fr ac{1}{4}(\pi + 2\gamma +6\ln 2 +2\ln\pi)$ . Remembering that $\sigma = (1-\sqrt{2})\zeta(\frac{1}{2})$ . Not sure yet, but this fact may shed light on whether $\sigma$ is rational or not, since this ratio is obviously irrational.

Because of the importance of the Riemann Zeta function, you will probably have more luck finding literature on the convergence of $\zeta(\frac{1}{2})$ than finding $\sigma$ directly.

Also, do you have any educated guesses on what $\sigma$ might be in terms of other known constants? It may easier to prove a correct-seeming hypothesis than to search for a quantity that may or may not exist.

**chisigma** · May 14th 2009, 05:35 AM

One of the reasons for which I have thanked Media_Man is what he reported from...

http://mathworld.wolfram.com/RiemannZetaFunction.html

... i.e. the value of $\zeta(\frac{1}{2})$ with [declared] 12 digits precision...

$\zeta(\frac{1}{2})= - 1.46035450881...$ (1)

This numerical value is taken from...

http://www.research.att.com/~njas/sequences/A059750

... where is shown with 99 digits precision!!

...

$\zeta(\frac{1}{2})= -1.4603545088095868128894$
$991525152980124672293310125814905428860878...$ (2)

All that gives me the opportunity to verify the precision of the result I have obtained working on an old Pentium PC to which I am attached for .... sentimental reasons

... but which is able to perform computations with only 12 digits accuracy

...

First step is now to prove the formula of Euler that I've 'discovered' ...

$\sum_{n=0}^{\infty} (-1)^{n} a_{n} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}} \sum_{k=0}^{n} (-1)^{k}\binom{n}{k} a_{n-k}$ (3)

... for some known result... for example the following...

$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} = \ln 2$ (4)

The calculator of Windows gives me $\ln 2$ with 32 digits accuracy...

$\ln 2 = .69314718055994530941723212145818\dots$

... that can be 'simplified' to 12 digit accuracy as...

$\ln 2 = .69314718056$ (5)

If we set in (3) ...

$a_{n} = \frac{1}{n+1}$

... and perform computation with the old Pentium we obtain...

$s(0) = .5$
$s(1) =$ $.625$
$s(2) =$ $.666666666667$
$s(3) =$ $.682291666667$
$s(4) =$ $.688541666667$
$s(5) =$ $.691145833333$
$\dots$
$s(34) =$ $.693147180559$
$s(35) =$ $.69314718056$ (6)

From (6) it is evident that 12 digit accuracy is obtained after 36 steps... not bad!

...

If we set in (3)...

$a_{n}=\frac{1}{\sqrt{n+1}}$ (7)

... and perform computation with the old Pentium we obtain...

$s(0) = .5$
$s(1) =$ $.573223304703$
$s(2) =$ $.593615393055$
$s(3) =$ $.600536047056$
$s(4) =$ $.60312629974$
$s(5) =$ $.604578400986$
$\dots$
$s(34) =$ $.604898643421$
$s(35) =$ $.604898643422$ (8)

For a singular coincidence also in this case 12 digits accuracy is obtained after 36 steps!... so that with 12 digits accuracy is...

$\sigma= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+1}}= .604898643422\dots$ (9)

According to...

http://en.wikipedia.org/wiki/Riemann_zeta_function

... was conjectured by Euler [always him and only him!!!

…] in 1749 that …

$\zeta(s)\cdot (1-2^{1-s}) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}$ (10)

... very well!!!... now if we take into account (9) and (10) and perform computation with the old Pentium we obtain with 12 digits accuracy...

$\zeta(\frac{1}{2}) = \frac{\sigma}{1- \sqrt{2}} = -1.46035450881...$ (11)

... i.e. the same as (1)... good old Pentium!

...

Kind regards

$\chi$ $\sigma$

**Media_Man** · May 14th 2009, 06:58 AM

In my previous posts I have been throwing a bunch of analytical stuff at the problem, instead of attacking the problem a priori. And here's some more (not for the faint)...

I uncovered $\eta'(s)=2^{1-s}\ln(2) \zeta(s)+(1-2^{1-s}) \zeta'(s)$ -----(1)

Letting $s=\frac{1}{2}$ , we get...

$\eta'(\frac{1}{2})=\sqrt{2}\ln(2) \zeta(\frac{1}{2})+(1-\sqrt{2}) \zeta'(\frac{1}{2})$ -----(2)

Dividing by $\zeta(\frac{1}{2})$ gives...

$\frac{\eta'(\frac{1}{2})}{\zeta(\frac{1}{2})}=\sqr t{2}\ln(2)+ (1-\sqrt{2})\frac{\zeta'(\frac{1}{2})}{\zeta(\frac{1} {2})}$ -----(3)

Substituting via $\eta(s)=(1-2^{1-s})\zeta(s)$ , we have...

$\frac{\eta'(\frac{1}{2})}{\eta(\frac{1}{2})}=\frac {\sqrt{2}\ln(2)}{(1-\sqrt{2})}+ \frac{\zeta'(\frac{1}{2})}{\zeta(\frac{1}{2})}$ -----(4)

And finally, utilizing $\frac{\zeta'(\frac{1}{2})}{\zeta(\frac{1}{2})}=\fr ac{1}{4}(\pi + 2\gamma +6\ln 2 +2\ln\pi)$ leaves us with the ratio:

$\frac{\eta'(\frac{1}{2})}{\eta(\frac{1}{2})}=\frac {\sqrt{2}\ln(2)}{(1-\sqrt{2})}+ \frac{1}{4}(\pi + 2\gamma +6\ln 2 +2\ln\pi)$ -----(5)

If your head is not yet spinning, you can verify the derivative:

Since $\eta(s)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s}$ ,
$\eta'(s)=\sum_{k=1}^\infty \frac{(-1)^k}{k^s}\ln(k)$

Obviously we cannot directly divide these two series to obtain an expression for the LHS of (5), but we do now have a second line of attack, since finding either $\eta(\frac{1}{2})$ OR $\eta'(\frac{1}{2})$ will allow us to find the other, since we know their ratio.

I think it is highly unlikely that either one of these is rational, but since the RHS of (5) is irrational, we know that they both cannot be rational. I will continue following this line of thought.

Thread: About an alternating sign series...

LinkBack

Thread Tools

Display

About an alternating sign series...

The Riemann Zeta Function

Known Results

Slightly more relevant...

Similar Math Help Forum Discussions

Alternating series

alternating series 1

alternating series #3

alternating series #2

alternating series

Search Tags