Proof that limit of sum of series is Omega needed

**Evo Rob** · July 12th 2008, 06:33 AM

Consider the following sum s(k), of k terms:

s(k) = sum for i = 1 to k of 1/((kW/(1-W)) + i -1)
where k and i are integers, and W is the Omega constant defined by ln(1/W) = W (W = 0.567143)

As an example, for k = 3, we have:
s(3) = 1/(3W/(1-W)) + 1/((3W/(1-W)) +1) + 1/((3W/(1-W)) +2)

Numerical analysis shows that the limit of this sum as k tends to infinity is W (the Omega constant). I am interested in finding a step-by-step analytical proof that this is the case.

A nice proof, or simply useful suggestions on how to attack the problem would be much appreciated.

Thanks

Rob

**flyingsquirrel** · July 12th 2008, 06:55 AM

Hi

Originally Posted by Evo Rob

Consider the following sum s(k), of k terms:

s(k) = sum for i = 1 to k of 1/((kW/(1-W)) + i -1)
where k and i are integers, and W is the Omega constant defined by ln(1/W) = W (W = 0.567143)

As an example, for k = 3, we have:
s(3) = 1/(3W/(1-W)) + 1/((3W/(1-W)) +1) + 1/((3W/(1-W)) +2)

Numerical analysis shows that the limit of this sum as k tends to infinity is W (the Omega constant). I am interested in finding a step-by-step analytical proof that this is the case.

A nice proof, or simply useful suggestions on how to attack the problem would be much appreciated.

$s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1}$ is almost a Riemann sum.

**CaptainBlack** · July 12th 2008, 10:51 AM

Originally Posted by flyingsquirrel

Hi

$s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1}$ is almost a Riemann sum.

Your suggestion does work:

$s(n)=\sum_{k=1}^n\frac{1}{\frac{n w}{1-w}+k-1} =\sum_{k=1}^n\frac{1}{n}\left[\frac{1}{\frac {w}{1-w}+(k-1)/n}\right]$

So $\lim_{n \to \infty} s(n)=\int_0^1 \frac{1}{\frac{w}{1-w}+x}~dx=\ln(1/w)=w$

(or was it a hint rather than a suggestion???)

Note we need not worry about if the sum should start from $0$ or $1$ or end at $n$ or $n-1$ , because in the limit it makes no difference.

RonL

**CaptainBlack** · July 12th 2008, 10:54 AM

Originally Posted by flyingsquirrel

Hi

$s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1}$ is almost a Riemann sum.

This is a hint, is it not???

RonL

**flyingsquirrel** · July 12th 2008, 11:08 AM

Originally Posted by CaptainBlack

This is a hint, is it not???

Yes, I meant that the limit of the sum can be computed using a Riemann sum.

**Evo Rob** · July 12th 2008, 11:52 AM

Thanks for the hint! After looking up what a Riemann sum is, and realising that the trick is to find the value of an equivalent integral, I have.

$ s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1} $

is a Riemann sum of the function f(x) = 1/x for the interval from
$ {\frac{n\Omega}{1-\Omega}} $ to $ {\frac{n\Omega}{1-\Omega}}+n $
Now $ \int_{b}^{c} 1/x = ln(c/b) $

Using integration to determine the limit of the sum, we get
$ ln({\frac{{\frac{n\Omega}{1-\Omega}}+n}{{\frac{n\Omega}{1-\Omega}}}}) $
$ =ln({\frac{1}{\Omega}}) = \Omega $

Thanks

Rob

**Mathstud28** · July 12th 2008, 12:14 PM

Originally Posted by Evo Rob

Thanks for the hint! After looking up what a Riemann sum is, and realising that the trick is to find the value of an equivalent integral, I have.

$ s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1} $

is a Riemann sum of the function f(x) = 1/x for the interval from
$ {\frac{n\Omega}{1-\Omega}} $ to $ {\frac{n\Omega}{1-\Omega}}+n $
Now $ \int_{b}^{c} 1/x = ln(c/b) $

Using integration to determine the limit of the sum, we get
$ ln({\frac{{\frac{n\Omega}{1-\Omega}}+n}{{\frac{n\Omega}{1-\Omega}}}}) $
$ =ln({\frac{1}{\Omega}}) = \Omega{\color{red}\leftarrow\text{???}} $

Thanks

Rob

...Since when has this been true

$\ln\left(\frac{1}{\Omega}\right)=\bold{\bold{\colo r{red}-\ln\left(\Omega\right)}}$

**Evo Rob** · July 12th 2008, 12:45 PM

My understanding is that the definition of $\Omega$ is

$ \ln(\frac{1}{\Omega})=\Omega $

Hence

$\Omega = 0.5671432904097838729999686622$

$ \ln(\frac{1}{\Omega})=-ln(\Omega) $

is also correct!

Rob

**flyingsquirrel** · July 12th 2008, 12:51 PM

Originally Posted by Evo Rob

Thanks for the hint! After looking up what a Riemann sum is, and realising that the trick is to find the value of an equivalent integral, I have.

$ s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1} $

is a Riemann sum of the function f(x) = 1/x for the interval from
$ {\frac{n\Omega}{1-\Omega}} $ to $ {\frac{n\Omega}{1-\Omega}}+n $

This interval shouldn't depend on $n$ .

$ s(n)=\sum_{k=1}^n\frac{1}{\frac{n\Omega}{1-\Omega}+k-1}=\frac{1}{n}\sum_{k=1}^n\frac{1}{\frac{\Omega}{1-\Omega}+\frac{k-1}{n}} $

and we are looking for something like $\sum_{k=1}^p (x_{k+1}-x_k)f(x_k)$

At this point we see that the function involved in the Riemann sum is $f:x\mapsto \frac{1}{x}$ , that the interval is $\left[ \frac{\Omega}{1-\Omega},\frac{\Omega}{1-\Omega}+1\right]$ , that the partition of this interval is made of the points $x_k=\frac{\Omega}{1-\Omega}+\frac{k-1}{n}$ for $1\leq k\leq n$ and that the distance between two consecutive points is $x_{k+1}-x_k=\frac{1}{n}$ . However, the last term of the partition has to be $\frac{\Omega}{1-\Omega}+1$ hence we add and subtract one term :

$\begin{aligned} s(n)&=-\frac{1}{n}\frac{1}{\frac{\Omega}{1-\Omega}+\frac{n}{n}}+\frac{1}{n}\frac{1}{\frac{\Om ega}{1-\Omega}+\frac{n}{n}}+\frac{1}{n}\sum_{k=1}^n\frac{ 1}{\frac{\Omega}{1-\Omega}+\frac{k-1}{n}}\\ &=-\frac{1}{n}\frac{1}{\frac{\Omega}{1-\Omega}+1}+\frac{1}{n}\sum_{k=1}^{n+1}\frac{1}{\fr ac{\Omega}{1-\Omega}+\frac{k-1}{n}}\\ \end{aligned} $
$ s(n)=-\frac{1}{n}\frac{1}{\frac{\Omega}{1-\Omega}+1}+\frac{1}{n}\sum_{k=0}^n\frac{1}{\frac{\ Omega}{1-\Omega}+\frac{k}{n}}\\ $

Now $\frac{1}{n}\sum_{k=0}^n\frac{1}{\frac{\Omega}{1-\Omega}+\frac{k}{n}}$ is a "true" Riemann sum which converges to $\int_{\frac{\Omega}{1-\Omega}}^{\frac{\Omega}{1-\Omega}+1}\frac{1}{x}\,\mathrm{d}x$ and the conclusion follows.

**Evo Rob** · July 12th 2008, 01:46 PM

However, the last term of the partition has to be

hence we add and subtract one term

Please excuse my ignorance, but is this adjustment necessary? The final "true" Riemann series appears to be for k = 0 to n, which is n+1 points. Is this correct?

I was working on the premise of a left Riemann series with n points. Is it ok to say that in the limit, this series is equivalent to the integral, without making the above adjustment?

Thanks

Rob

**flyingsquirrel** · July 12th 2008, 02:22 PM

Originally Posted by Evo Rob

Please excuse my ignorance, but is this adjustment necessary? The final "true" Riemann series appears to be for k = 0 to n, which is n+1 points. Is this correct?

I was working on the premise of a left Riemann series with n points. Is it ok to say that in the limit, this series is equivalent to the integral, without making the above adjustment?

Yes, it's OK. The extra term I added/subtracted is useless, the "true" Riemann sum is not a Riemann sum... you can directly conclude from $s(n)=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{\Omeg a}{1-\Omega}+\frac{k-1}{n}}$ as you want to do it. My mistake, sorry.

**Evo Rob** · July 13th 2008, 01:48 AM

flyingsquirrel, CaptainBlack, Mathstud28,

Thank you for your contributions to this thread, and for checking my maths. I am pleased to have found a proof and to have worked it out, for the most part, myself from the helpful hint.

Cheers

Rob

**CaptainBlack** · July 13th 2008, 01:59 AM

Originally Posted by Evo Rob

My understanding is that the definition of $\Omega$ is

$ \ln(\frac{1}{\Omega})=\Omega $

Hence

$\Omega = 0.5671432904097838729999686622$

$ \ln(\frac{1}{\Omega})=-ln(\Omega) $

is also correct!

Rob

You may have noticed that in my posts that I have use $w$ for this rather than $\Omega$ mainly because I suspect that the original source was using $w$ , since this is related to Lambert's W-function, where $W(z)$ satisfies the equation:

$z=W(z)e^{W(z)},$

so:

$\ln(1/W(1))=W(1)$

RonL

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