1

Earlier i've asked about how to calculate divergent products, i got some directions which made me curious. Now i'm wondering is this correctly done.

Divergent products.

The most commen divergent product, and also mentioned in the topic quite often was: $$\prod_{n=1}^{\infty}(n)= (2\pi)^{0.5} $$ So i wanted to check it myself.

I used the following formula.

$$\sum_{n=1}^{dp} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} (d f(nd)-f(n))$$

With d an integer (the period) and d>=2, and p tends to infinity.

I got this formula the following way:

$$\sum_{n=1}^{dp} \sum_{z=0}^{d-1}f(n+z)\sum_{k=0}^{d-1} (e^{\frac{2i\pi k}{d}})^{n+z}=\sum^p_{n=1} d^2 f(nd)$$

And after corrections, this gives : $$\sum_{n=1}^{dp} \sum_{z=0}^{d-1}f(n+z)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n+z}-\sum^{d-1}_{z=1}(d-z)f(z)=\sum^p_{n=1} (d^2 f(nd)-df(n))$$

Which give: $$\sum_{n=1}^{dp} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} (d f(nd)-f(n))$$

I used this also to calculate the zeta funtions for example. There are some function were this won't work, or need to be correct because of constants, so question 1, what is the "real" mathematical formula to transform (divergent) sums into it's alternating form.

$$\prod_{n=1}^{\infty}(n)= e^{\sum_{n=1}^{\infty} ln(n)}$$ $$ \sum_{n=1}^{\infty} ln(n)= \sum_{n=1}^{\infty} 2ln(2n/4)-ln(1/4) =\sum_{n=1}^{\infty} (-1)^n*ln(n/4)$$

$$\sum_{n=1}^{\infty} -(-1)^n*ln(4)=1/2 (ln(4))$$

So i'm left with $$\sum_{n=1}^{\infty} (-1)^n*ln(n)= $$ $$1/2(\sum_{n=1}^{1/2(m)} ln(2n)-\sum_{n=1}^{1/2(m)} ln(2n-1) + \sum_{n=1}^{1/2(m-1)} ln(2n) -\sum_{n=1}^{1/2(m+1)} ln(2n-1) )=$$ $$ (2m-1)/2*ln(2)+ln((1/2*m)!)+ln((1/2(m-1))!)-ln(m!)=$$ $$ -1/2ln(2)+ln(\frac{((1/2)m)!*((1/2*(m-1))!)*2^m}{m!})=1/2(ln(pi/2))$$

Which solved the problem: $$\prod_{n=1}^{\infty}(n)= (2\pi)^{0.5} $$

Because i'm no mathematician, everything i did is done on intuition, is this correctly done? And i have troubles solving the following sum, if i want to use the same idea, how should i solve: $$ \sum_{n=1}^{\infty} ln(n+1) $$.

Community
  • 1
Gerben
  • 593
  • The product of interest does not converge. – Mark Viola Jun 14 '15 at 16:57
  • 4
    It's why i said divergent products and sums in the title. – Gerben Jun 14 '15 at 18:16
  • 1
    Why did you write that the divergent product equals $\sqrt{2\pi}$? – Mark Viola Jun 14 '15 at 18:27
  • 3
    Why i did write it down in the first place, was that it was stated in the previous topic as derivative value of zeta(o). Since it's hard, atleast for me, to find the derivative of the zeta function, i tried if there was an easier way to calculate it. So when i did it myself (as shown above) i get the same value. Why i assign a finite value to a divergent product is because i belief it's the constant finite part. – Gerben Jun 14 '15 at 19:16
  • @MarkViola $\prod_n n = \exp \sum_n n^{-s} \log n |{s = 0} = \exp \left(-\sum_n \frac{d}{ds} n^{-s} |{s = 0}\right) = \exp \left(-\zeta'(s)|_{s=0}\right) = \exp \left(-\zeta'(0)\right) = \sqrt{2\pi}$. – user76284 Mar 05 '20 at 19:46
  • @user76284 You're trying to regularize the divergent product. The manipulations you've written are completely non-rigorous. – Mark Viola Mar 05 '20 at 21:50
  • @MarkViola "You're trying to regularize the divergent product." That's exactly what the question is about? – user76284 Mar 05 '20 at 21:52
  • @user76284 "That's exactly what the question is about?" Are you asking a question? And yet you're formal manipulations lack rigor. – Mark Viola Mar 06 '20 at 03:21
  • @MarkViola The question is about regularizing divergent products. – user76284 Mar 06 '20 at 03:23

0 Answers0