# 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.

• In any case, this is beautiful ! – Claude Leibovici Mar 5 '14 at 13:32
• :-) Thanks, @Claude! $\phantom{....................}$ – Gottfried Helms Mar 5 '14 at 19:24
• Jesus, these are merely your "exercises"? – Simply Beautiful Art Dec 4 '16 at 18:21
• @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