Recently i've thought of a new step in my summations. I've been mentioning them in my previous questions. And i know there are certain point were it doesn't work. But most of the time it does.
$$\sum_{n=1}^{dp} \sum_{z=0}^{d-1}f(n+z)\sum_{k=0}^{d-1} (e^{\frac{2i\pi k}{d}})^{n+z}=\sum^p_{n=1} d^2 f(nd)$$ $$\sum_{n=1}^{dp} \sum_{z=0}^{d-1}f(n+z)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n+z}-\sum^{d-1}_{z=1}(d-z)f(z)=\sum^p_{n=1} (d^2 f(nd)-df(n))$$ $$\sum_{n=1}^{dp} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} (d f(nd)-f(n))$$ $$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty}\sum_{p=1}^{d-1} -(d)^{k}f(d^kn)e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}f(n) $$
Until here i found it to working perfectly fine with quite some examples of cool, for me, new formula's. Question: are the following steps correct.
$$\sum_{d=2}^{m}\sum_{k=0}^{\infty} \sum_{n=1}^{\infty}\sum_{p=1}^{d-1} -(d)^{k}f(d^kn)e^{2i\pi*pn/d}=(m-1)\sum_{n=1}^{\infty}f(n) $$
The downarrow means rounding down; ncd means no common divisors of (d). So ncd(10) are 1,3,7,9. $$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{m/d\downarrow}\sum_{p=ncd(d)}^{d-1}\sum_{k=0}^{\infty} (sd)^{k}f((sd)^k)n)e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}f(n) $$
Example: from here on it's for every step $\lim_{m\to\infty} $ $$f(n)=1/n^2$$
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{m/d\downarrow}\sum_{p=ncd(d)}^{d-1}\sum_{k=0}^{\infty} \frac{1}{(sd)^{k}n^2} e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}1/n^2 $$
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{m/d\downarrow}\sum_{p=ncd(d)}^{d-1} \frac{sd}{1-sd}1/n^2 e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}1/n^2 $$ $$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{p=ncd(d)}^{d-1} -m/d*1/n^2 e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}1/n^2 $$ $$\sum_{n=1}^{\infty}\sum_{d=2}^{\infty}\sum_{p=ncd(d)}^{d-1} \frac{e^{2i\pi*pn/d}}{dn^2}=\sum_{n=1}^{\infty}1/n^2 $$
\begin{array}{|c|c|c|} \hline d & value & sum \\ \hline 2 & 0041666667, & 0,041666667 \\ 3 & 0,037037037 & 0,078703704 \\ 4 & 0,010416667 & 0,08912037 \\ 5 & 0,033333333 & 0,122453704 \\ 6 & -0,009259259 & 0,113194444 \\ 7 & 0,020408163 & 0,133602608 \\ 8 & 0,002604167 & 0,136206774 \\ 9 & 0,004115226 & 0,140322001 \\ 10 & -0,006666667 & 0,133655334 \\ 11 & 0,013774105 & 0,147429439 \\ 12 & -0,002314815 & 0,145114624 \\ 13 & 0,01183432 & 0,156948943 \\ 14 & -0,005102041 & 0,151846903 \\ 15 & -0,005925926 & 0,145920977 \\ 16 & 0,000651042 & 0,146572018 \\ \hline \end{array}
For me, all of this is bloody hard to check. So i'm afraid of giving more general exmaples of things ivé thought out. I'm used to work with only excel and paper. In this tabel i've dealt $\pi^2$ out of it. Question, i think this goes to $1/6$ since its the giving sum, is this correct?
Edit: working on my second more advanced example of the divergent sum $\zeta(1)$: The divergent series of f(n)=n would be the following. I haven't been unable to check this, so i might be totaly wrong.
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{m/d\downarrow}\sum_{p=ncd(d)}^{d-1}\sum_{k=0}^{\infty} n(sd)^{2k} e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}n $$
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{m/d\downarrow}\sum_{p=ncd(d)}^{d-1} \frac{n}{(sd)^2-1} e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}n $$
Since the sum over s is convergent
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{s=1}^{\infty}\sum_{p=ncd(d)}^{d-1} \frac{n}{(sd)^2-1} e^{2i\pi*pn/d}=\sum_{n=1}^{\infty}n $$
$$\frac{1}{1-m}\sum_{n=1}^{\infty}\sum_{d=2}^{m}\sum_{p=ncd(d)}^{d-1}(\frac{1}{2}-\frac{\pi}{2d\tan(\pi/d)})ne^{2i\pi*pn/d}=\sum_{n=1}^{\infty}n $$
$$\frac{1}{1-m}\sum_{d=2}^{m}\sum_{p=ncd(d)}^{d-1}(\frac{1}{2}-\frac{\pi}{2d\tan(\pi/d)})\frac{e^{2i\pi*p/d}}{(e^{2i\pi*p/d}-1)^2}=\sum_{n=1}^{\infty}n $$
I"m unable to compute this myself, but does this actualy go to $-1/12$ ( or 1/12 if i missed a minus sign).