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.
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: $$ \sum_{n=1}^{\infty} ln(n+1) $$.
