# A Ramanujan-like summation: is it correct? Is it extensible?

I'm still exercising with summation-procedures which I try to make correct Ramanujan-summations. I'm looking at the (gap-)series $$s(1/2,2) = (1/2)^1+(1/2)^{4}+(1/2)^{9}+(1/2)^{16}+(1/2)^{25}+...$$ and more general at $$s(b,p) = b+b^{2^p}+b^{3^p}+b^{4^p}+b^{5^p}+... \tag 1$$ For convergent case where $0<b<1$ we can evaluate this approximately using serial evaluation of expression (1), say $$\begin{eqnarray} s(1/2,1) &=& 1 \\s(1/2,2) & =& 0.564468413... \\ s(1/2,3)&=&0.503906257... \end{eqnarray}$$.

On the other hand, expanding that formal sum in a double series of a series of formal exponential series in $x$ and collecting like powers of the argument $x$ I got something like Ramanujan-summation (where I only express the Bernoulli-numbers in the Ramanujan formula by the equivalent zeta-references at negative arguments).
So I define the expressions $$\begin{eqnarray} I(b,p)&=& \int_{t=- 1}^\infty b^{(1+t)^p} dt \\ Z(b,p)&=& \sum_{k=0}^\infty \beta^k {\zeta(-pk) \over k!} \qquad \text{ where } \beta = \log(b) \end{eqnarray} \tag 2$$ Then the summation-method $\mathcal Q$ $$s(b,p) \underset{\mathcal Q}= I(b,p) + Z(b,p) \tag 3$$ gives for the above convergent cases $s(1/2,1),s(1/2,2)$ the correct results by numerical approximations. (The second case is convergent because each zeta at negative even argument is zero). $$\begin{eqnarray} I(1/2,1) &=& + 1.44269504089... \\ Z(1/2,1) &=& -0.44269504089...\\ s(1/2,1) &\underset{\mathcal Q}=& \phantom + 1\\ \hline\\ I(1/2,2) & =& + 1.06446701943... \\ Z(1/2,2)&=& -0.5\\ s(1/2,2)&\underset{\mathcal Q}=& \phantom + 0.56446701943... \end{eqnarray}$$.

Also various examples with other parameters, where everything in the $\mathcal Q$-method is convergent seems to approximate the results taken by serial evaluation of (1) correctly. For the case $s(1/2,3)$ the sum $Z(1/2,3)$ in (2) is no more convergent. However, using a Noerlund-summation $\mathcal N$ for that expression I could approximate the expected result (taken by evaluation of (1)) satisfyingly. $$\begin{eqnarray} I(1/2,3) & =& + 1.00901976692... \\ Z(1/2,3)&\underset{\mathcal N}=& -0.50561...\\ s(1/2,3)&\underset{\mathcal Q}=& \phantom + 0.50340...\\ \hline \\ s(1/2,3)&=& \phantom + 0.503906257 \text{ by serial summation } \end{eqnarray}$$.

So perhaps this allows more generalization.

Q1: Is this eq (2) a valid reconstruction of the Ramanujan-summation, at least in principle?

Q2: The integration-bounds were experimentally. Are they correct? and if: how could I have derived them correctly?

[update1]: There must be some systematical error. For all even $p=2q$ the sum of the zetas $Z(1/2,2 \cdot q) =-1/2 = \zeta(0)$ and the correct sum $s(1/2,2\cdot q) = 1/2 + \delta$ is always bigger than $1/2$. So it is required that $$I(1/2,2\cdot q)= s(1/2,2\cdot q) - Z(1/2,2 \cdot q) = 1 +\delta$$
But now, by Wolframalpha I see that $$\int_{t=-1}^\infty 1/2^{(1+t)^p} dt = { \Gamma(1+1/p)\over \sqrt[p]{\log 2} }$$ which decreases below $1$ at $p \approx 3.44395$ so for all $p=4,6,8,...$ the $\mathcal Q$-summation cannot hold.
Hmm...

[update 2,update 3] Here are some data which show very nice approximation for the $\mathcal Q$-summation with the serial summation for small p of $p=0.1 \ldots 1.2$ with errors $\lt 1e-100$. $$\small \begin{array} {r|rrl} p & s(1/2,p) & err &=s(1/2,p) - (I(1/2,p)+Z(1/2,p) \\ \hline 0.1 & 141745219.752 & -4.00E-156 \\ 0.2 & 749.678969635 & 2.82E-167 \\ 0.3 & 31.0876588123 & 4.18E-166 \\ 0.4 & 7.95318949339 & 1.33E-164 \\ 0.5 & 3.78821923065 & -1.96E-162 \\ 0.6 & 2.37977624581 & 9.16E-160 \\ 0.7 & 1.72998968339 & -2.28E-156 \\ 0.8 & 1.37118947539 & -6.79E-152 \\ 0.9 & 1.14893798357 & -4.14E-146 \\ 1.0 & 1.00000000000 & 5.69E-139 \\ 1.1 & 0.89438296309 & -1.18E-128 \\ 1.2 & 0.81625996055 & 2.25E-114 \\ &&& \text{very small errors - perhaps is \mathcal Q-summation valid? }\\ \vdots \\ 1.7 & 0.61712884341 & 2.28 E-18 & \text{using Borel-summation}\\ 1.8 & 0.59639864943 & -1.27 E-11 \\ 1.9 & 0.57905393685 & -0.0000000385 \\ &&& \text{errors increase- \mathcal Q-summation tends to become invalid? }\\ \vdots\\ 3.0 & 0.503906257 \phantom {00} & 0.00040176 & \text{Z(1/2,p) by Noerlund/Borel-summation } \end{array}$$ Because of the very slow convergence at the small p I used the Pari/GP-function sumpos which seems to be able to approximate the true value avoiding the computation of millions of terms in the serial summation (1). For the parameter $p \gt 1.5$ I crosschecked the Noerlund-summation for $Z(1/2,p)$ by Borel-summation.

• @SimpleArt: well, I'm not a professional, so what should I say instead of "exercises"... ;-) "Ramanujan-summation" is not easy to understand and it seems it is better to get a grip by various example-attempts. On the other hand, the software Pari/GP allows to reduce experimental, numerical mathematics sometimes to procedures, which can be, well, handled. - In the end: english is not my native language, so I hope I didn't miss some (possible) cheek-in-tongue? – Gottfried Helms Dec 4 '16 at 18:49
• Not sure how helpful this is, but you can probably write $s(b,p)$ in terms of Jacobi theta functions. – Simply Beautiful Art Sep 22 '17 at 0:20
• Have you tried a Similair approach as sum e^(-n^2)? Use sum c^(n^2)= sum ln(c)^(m/2) zeta(-m)/(m/2)! (1+(-1)^m)/2. Find the poles, at m=-1 and i think only at m=0 are only needed for the even m. Reconise the zeros at uneven are growing if m goes to either infinity, rewrite the sum with reflections formulas to find a regularized solution. Guess Monday is going to be a busy evening ;) . I will post a try myself later on, currently not behind a pc, but it seems doable just the "regular" way. – Gerben Oct 10 '20 at 10:50

A non rigorous approach, just simple "intuitive methode". Because the power is $$x^2$$, the period is 2. Make sure to include the zero point in it's polynomal notation. I will later try to write it for all power's of $$x$$.

$$\sum_{x=1}^{\infty} (c)^{-x^2}$$ $$\sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ (\ln(c)x^2)^n(-1)^n/n!=\sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ \frac{(x)^n (\ln(c))^{n/2}}{(n/2)!} (e^{i\pi n/2})\sum_{k=0}^1 \frac{e^{\frac{2i \pi*kn}{2}}}{2}$$

$$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} \frac{(e^{i\pi n/2}+e^{3i\pi n/2})}{2}$$

First take the limit as n goes to -1. Which gives us the value $$\frac{ \sqrt{\pi}}{2 \sqrt{\ln(c)}}$$

There's another value at the "normal" order, mainly as $$n$$ goes to 0. At $$n=0$$ the value is $$-\frac{1}{2}$$.

Notice that both answers give a very very good approaximation for the sum, but are a very little off, (orders like $$10^-{6}$$, really small). Mathematically I can't fully show why (yet). More on intuition based, let $$h$$ be a very small value, and you have rewrite the function you are trying to sum as $$g(n)f(n)$$ with $$g(n)$$ being a periodically zero, with an order $$h$$ near the zero so $$g(n+h)$$ and $$f(n)$$ being a function that grows more/equal then a constant value, the regularised (as in the upper or lower limit is infinity) the sum is not that well defined and you just have to be careful. Lots of horrible talk and explainations I soon will be ashamed of so I am hoping for a better explaination. So here's what I found to be the solution.

If $$n$$ is even, the zero's don't have any influence for the sum and can be regarded as 0.

If n is uneven this isn't the case: $$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} e^{i\pi n/2} \frac{(1+(-1)^n)}{2}$$

use that $$\zeta(-n)=\frac{\zeta(n+1) n! (i^{n+1}+(-i)^{n+1})}{(2\pi)^{n+1}}$$ $$\frac{n!}{(n/2)!2^n}=\frac{((n-1)/2)!}{\sqrt{\pi}}$$

$$\sum_{n \in \mathbb{Z}}\ \frac{\zeta(n+1)(\ln(c))^{n/2}\big(\frac{(n-1)}{2}\big)! }{\pi^{n+3/2}} \frac{(e^{i\pi n/2}+e^{3i\pi n/2}))((e^{i\pi (n+1)/2}+e^{3i\pi (n+1)/2}))}{4}=$$

We only had to look at when $$n=2m-1$$.

$$\sum_{m \in \mathbb{Z}}\ \frac{\zeta(2m)(m-1)! (ln(c))^{m-1/2} }{2\pi^{2m+1/2}} i(e^{3i\pi m}-e^{i\pi (m)})(-1)^{m}$$ Because it's summed over all integers including the negative ones, we subsitute m=-m.

$$\sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m-1/2}}{2(\ln(c))^{m+1/2}} i(e^{-3i\pi m}-e^{-i\pi (m)}) (-1)^{m}$$

I've had the alternating/peridiocally zero's kinda horrible defined. But approach the limit of m=m+h at the zero's/infinities to look at the (uneven) zero's. So that $$e^{-3i\pi (m+h)}-e^{-i\pi (m+h)}=-2hi\pi$$ I know you should work this better out with it's period, and notice that everything is defined in the same "complex" period, maybe use trigonometry functions, but those functions are less intunitive. Again it's non rigorous, only to show that there's some way to a good solution with this methode. And know that $$(-n-1)!=h^{-1}/n!$$ if n is an integer $$\sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m+1/2}}{(\ln(c))^{m+1/2}} (h) (-1)^{m}$$ $$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-2m) \pi^{2m}}{m!(\ln(c))^{m}} (-1)^{m}$$ Officially I'd say you'd had to write it again out also as (and maybe you couldn't ignore the other zero's, I just don't know yet, it just works, I've to polish it obviously)

$$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-m) \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ $$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}\sum_{m \in \mathbb{Z}} \frac{x^{m} \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ Which is the regularised sum of $$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{ln(c)}}$$

$$\sum_{x=1}^{\infty} (c)^{-x^2}=\frac{\sqrt{\pi}}{2 \sqrt{ln(c)}}-1/2+\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{\ln(c)}}.$$

The first two value's being your I(c^(-1),2) and Z(c^(-1),2) and the sum being your systematically error. And i can only agree that the systematically error is something counter intunitive at first, but also arise when you apply this summation methode to other series as i recall well known easier examples such as $$\sum_{x=1}^{\infty} \frac{1}{2x^2-1}$$.

In similair rough fashion for $$\sum_{x=1}^{\infty} (c)^{-x^d}$$

$$\sum_{m \in \mathbb{Z}}\ \frac{(-1)^{1/d}\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \sum_{k=0}^{d-1} \frac{e^{\frac{2ipi*km}{d}}}{d}$$

$$\sum_{m \in \mathbb{Z}}\ \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

Now the value at m=0, will always be -1/2. at limit m goes to -1 gives:

$$\frac{-\pi}{\sin(\pi/d) (\ln(c))^{1/d}(-\frac{1+d}{d})!}=\frac{(1/d)!}{\ln(c)^{1/d}}$$

Which is similair to the integeral you stated, if d is not even you got to indeed sum is over the other value's and don't forget the error.

A try of mine to find the error:

$$\sum_{m \in \mathbb{Z}} \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{h \pi}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m+1} (m/d)!} \frac{h \pi (i^{m+1}+(-i)^{m+1})}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m} (m/d)!} \frac{h \pi (sin(m \pi/2)}{d \sin(m \pi /d)}$$

Use that:

$$\frac{m!}{(m/d)!}= d^{m+1/2} (2 \pi)^{(1-d)/2} \prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) (ln(c))^{m/d}}{} \frac{h (sin(m \pi/2)}{\sin(m \pi /d)}\frac{d^{m-1/2} (2 )^{(3-d)/2-m} (\pi)^{(1-d)/2-m}\prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!}{}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(1-d)/2+m}\prod_{j=1}^{d-1} \big(\frac{-m-j}{d}\big)!}{}$$

Split up for the zero's in $$\prod_{j=1}^{d-1}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(3-d)/2+m}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (m+j)}{d}) \big(\frac{m+j}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{} \frac{h (sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(3-d)/2+dm+k}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{ \pi (dm+j+k)}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{(sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})=d2^{1-d}(-1)^{?}$$ I don't have the minus sign for now. Will come back at that later, assuming it's not there cause i've already enough in one equation.

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{}\frac{d^{(-dm-k)-3/2} (2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

to make it readable $$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}\frac{(2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{d^{(dm+k)+3/2} }$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} \frac{(\frac{2\pi}{d})^{dm+k+1/2} (\frac{2}{\pi})^{d/2}}{d}$$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

I am already suprised the formula above seem to work for d=2. If only the m didn't sneaked into the product life would be good. Kinda used this is a note pad, i will clean it up one day, but results count.

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(m)}}{} \frac{\sin((m) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{m+j}{d}-1\big)!}$$

It shows how I think you can proceed (I let it rest, as it got messy, and i probably made some errors, and made things unnessecary hard), with a doable solution of d=2, as was the original question. Maybe you can or should add $$m \in d\mathbb{Z}$$, to make it smooth, Again i do not know yet.