Sum Over n

Main

We can package a sum over \(n\) into a set of sums with a triangular number like stride. We can see this by taking the integers, removing the triangular numbers, then shifting by one, removing the triangular numbers again

\begin{equation} 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,\cdots\\ .,2,.,4,5,.,7,8,9,.,11,12,13,14,.,16,17,18,19,20,.,22,23,24,25,26,27,.,29,30,\cdots\\ .,.,.,.,5,.,.,8,9,.,.,12,13,14,.,.,17,18,19,20,.,.,23,24,25,26,27,.,.,30,\cdots\\ .,.,.,.,.,.,.,.,9,.,.,.,13,14,.,.,.,18,19,20,.,.,.,24,25,26,27,.,.,.,\cdots\\ \\ \end{equation}

and so on

\begin{equation} \sum_{n=1}^{\infty}f(n)=\sum_{n\in\{1,3,6,10,15\}}f(n)+\sum_{n\in\{2,4,7,11,16\}}f(n)+\sum_{n\in\{5,8,12,17,23\}}f(n)+\cdots\\ \end{equation}

the first coefficients of the latte sums take the formula

\begin{equation} c_{1,0}=1\\ c_{1,n}=\frac{1}{2}(3n+n^{2})\\ \end{equation}

where we are calling the first sum \(0\) and the later sums \(1,2,3,4,\cdots,n,\cdots\). Then we have that

\begin{equation} c_{n,0}=\frac{n(n+1)}{2}\\ c_{n,1}=1+\frac{n(n+1)}{2}=1+c_{n,0}\\ c_{n,2}=2+\frac{(n+1)(n+2)}{2}=1+c_{n+1,1}=2+c_{n+1,0}\\ c_{n,3}=3+c_{n+2,0}\\ c_{n,4}=4+c_{n+3,0}\\ c_{n,m}=m+c_{n+m-1,0}=m+\frac{(n+m-1)(n+m)}{2},\;m>0\\ \end{equation}

so we can rewrite

\begin{equation} \sum_{n=1}^{\infty}f(n)=\sum_{n=1}^{\infty}f(c_{n,0})+\sum_{n=1}^{\infty}f(c_{n,1})+\sum_{n=1}^{\infty}f(c_{n,1})+\cdots\\ \sum_{n=1}^{\infty}f(n)=\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}f(c_{n,m})\\ \sum_{n=1}^{\infty}f(n)=\sum_{n=1}^{\infty}f\left(\frac{n(n+1)}{2}\right)+\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f\left(m+\frac{(n+m-1)(n+m)}{2}\right)\\ \end{equation}

Is this useful? It seems to complicate things alot, for example the normal example

\begin{equation} \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\\ \end{equation}

The first term is easy

\begin{equation} S_{0}=\sum_{n=1}^{\infty}\left(\frac{n(n+1)}{2}\right)^{-2}=\frac{4}{3}(\pi^{2}-9)\\ \end{equation}

the rest is then

\begin{equation} S_{1}=\sum_{n=1}^{\infty}\frac{4}{(1+8n)^{3/2}}\left(2(\psi_{0}(t_{-})-\psi_{0}(t_{+}))+\sqrt{1+8n}(\psi_{1}(t_{-})+\psi_{1}(t_{+}))\right)\\ \end{equation}

with \(\psi_{n}\) the polygamma function, and

\begin{equation} t_{+}=\frac{3}{2}+n+\frac{1}{2}\sqrt{1+8n}\\ t_{-}=\frac{3}{2}+n-\frac{1}{2}\sqrt{1+8n}\\ \end{equation}

we then must have that \(S_{0}+S_{1}=\pi^{2}/6\). So

\begin{equation} S_{1}=\frac{1}{6}(72-7\pi^{2})\\ \end{equation}

But we can also write a chain of expressions for the \(m^{th}\) sum,

\begin{equation} \frac{\pi^{2}}{6}=\frac{1}{3}(4\pi^{2}-36)+\frac{1}{27}(4\pi^{2}-31)+S_{2}+\cdots\\ \end{equation}

but some of these sums are hard to write here. We find the value indeed converges towards the required value of \(S_{1}\approx 0.485462\). The sum is not particularly efficient.

Zeta(3)

We can work towards a sum for \(\zeta(3)\) though! We find from similar treatment we have

\begin{equation} \zeta(3)=8(10-\pi^{2})-\sum_{n=1}^{\infty}\frac{4}{(1+8n)^{5/2}}\Bigg{(}12(\psi_{0}(t_{-})-\psi_{0}(t_{+}))+6\sqrt{1+8n}(\psi_{1}(t_{-})+\psi_{1}(t_{+}))+(1+8n)(\psi_{2}(t_{-})-\psi_{2}(t_{+}))\Bigg{)}\\ \end{equation}

and the remaining terms converge on the correct number \(0.15889211187406554\). For triangular numbers, the summand simplifies to a nicer form.

Sum

\begin{equation} \sum_{n=1}^{\infty}\frac{1}{2^{n}}=1\\ \end{equation}

so then when we perform

\begin{equation} \sum_{n=1}^{\infty}2^{-(n(n+1)/2)}=\frac{1}{2^{7/8}}\theta_{2}(0,\frac{1}{\sqrt{2}})-1\\ \end{equation}
\begin{equation} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}2^{-\left(m+\frac{(n+m-1)(n+m)}{2}\right)}\\ \end{equation}

for \(m=1\) we get

\begin{equation} \frac{1}{2\cdot 2^{7/8}}\theta_{2}(0,\frac{1}{\sqrt{2}})-\frac{1}{2}\\ \end{equation}

Generating Functions

We may write the set of powers of \(x\) as generating functions as follows

\begin{equation} T=\frac{1}{2}\theta_{2}(0,\sqrt{x})\\ \end{equation}

for elliptic theta, then

\begin{equation} -1+x^{-1/8}T=x+x^{3}+x^{6}+x^{10}+x^{15}+x^{21}+x^{28}+\cdots\\ -x+x^{7/8}T=x^{2}+x^{4}+x^{7}+x^{11}+x^{16}+x^{22}+\cdots\\ -x^{2}(1+x)+x^{15/8}T=x^{5}+x^{8}+x^{12}+x^{17}+x^{23}+x^{30}+\cdots\\ \\ \end{equation}

and the general term becomes

\begin{equation} -1+x^{-1/8}T,\;n=0\\ -x^{n}\left(\sum_{k=0}^{n-1}x^{k(k+1)/2}\right)+x^{(8n-1)/8}T=\sum_{k=1}^{\infty}x^{n+(k+n-1)(k+n)/2},\;n>0\\ \end{equation}

Thus

\begin{equation} \frac{x}{1-x}=-1+x^{-1/8}T+\sum_{n=1}^{\infty}-x^{n}\left(\sum_{k=0}^{n-1}x^{k(k+1)/2}\right)+x^{(8n-1)/8}T\\ \frac{x}{1-x}=A+\sum_{n=1}^{\infty}-x^{n}\left(\sum_{k=0}^{n-1}x^{k(k+1)/2}\right)\\ \end{equation}

where

\begin{equation} A=-1+x^{-1/8}T+\sum_{n=1}^{\infty}x^{(8n-1)/8}T\\ A=\frac{\theta_{2}(0,\sqrt{x})}{(2-2x)x^{1/8}}-1\\ \end{equation}

where \(A+1\) is the beautiful series

\begin{equation} 1+2x+2x^{2}+3x^{3}+3x^{4}+3x^{5}+4x^{6}+4x^{7}+4x^{8}+4x^{9}+\cdots\\ \end{equation}

with \(n\) of each coefficient! This then gives

\begin{equation} B=\sum_{n=1}^{\infty}-x^{n}\left(\sum_{k=0}^{n-1}x^{k(k+1)/2}\right)=-x-x^{2}-2x^{3}-2x^{4}-2x^{5}-3x^{6}-3x^{7}-3x^{8}-3x^{9}-\cdots\\ \end{equation}

Squares

We can repeat the above, but using square numbers instead of triangle numbers. A sum over the integers can be broken down into the following expressions

\begin{equation} n^{2}\\ n^{2}+1\\ n^{2}+2\\ (n+1)^{2}+3\\ (n+1)^{2}+4\\ (n+2)^{2}+5\\ ..\\ \end{equation}

Then

\begin{equation} \sum_{n}f(n)=\sum_{n=1}^{\infty}f(n^{2})+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}f\left((n+\lfloor\frac{m}{2}\rfloor)^{2}+m+1\right)\\ \end{equation}

We can take the divergent sum

\begin{equation} Q=\sum_{n=1}^{\infty}\frac{1}{n}\\ \end{equation}

and then go through this process and state that

\begin{equation} Q=\zeta(2)+\frac{1}{2}(\pi\coth(\pi)-1)+\frac{1}{4}(\sqrt{2}\pi\coth(\sqrt{2}\pi)-1)+\frac{1}{12}(2\sqrt{3}\pi\coth(\sqrt{3}\pi)-5)+\frac{1}{40}(5\sqrt{4}\pi\coth(\sqrt{4}\pi)-13)+\cdots\\ \end{equation}

The fractional part of one over the convergents appears to tend to a constant for the \(n^{th}\) term! (Worth investigating?) The constant is around \(0.165-0.168\). The non-fractional part seems to repeat the integers. It appears to be \(0.166666666...\), i.e a sixth.

Then for large \(m\), the \(m^{th}\) sum equals approximately

\begin{equation} \frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\ \end{equation}

for some term

\begin{equation} \frac{1}{a_{m}}\left(b_{m}\sqrt{m}\pi\coth(\sqrt{m}\pi)-c_{m}\right)\to\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\ \end{equation}

we know that for large \(m\) , \(coth(m)=1\), so then we have

\begin{equation} \lim_{m\to\infty}\frac{b_{m}\sqrt{m}\pi-c_{m}}{a_{m}}\to\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\ a_{m}=\{2,4,12,40,90,420,1232,2448,5850\cdots\}\\ b_{m}=\{1,1,2,5,9,35,88,153,325\cdots\}\\ c_{m}=\{1,1,5,13,34,137,431,773,1919,\cdots\}\\ \end{equation}

but we know that

\begin{equation} \sum_{m=0}^{\infty}\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\to\infty\\ \end{equation}

as does \(\zeta(1)\), which we were summing to begin with. It would appear by inspection that

\begin{equation} \frac{a_{m}}{2m}=\{1,1,2,5,9,35,88\}=b_{m}\\ \end{equation}

where \(b_{m}\) appears more basic than \(a_{m}\), so we can write

\begin{equation} \lim_{m\to\infty}\frac{b_{m}\sqrt{m}\pi-c_{m}}{2mb_{m}}\to\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\ \lim_{m\to\infty}\frac{\pi}{2\sqrt{m}}-\frac{c_{m}}{2mb_{m}}\to\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\ \\ \end{equation}

but really we had noticed that the fractional part tends to \(1/6\), so we should state that

\begin{equation} \lim_{m\to\infty}\frac{2mb_{m}}{\pi b_{m}\sqrt{m}-c_{m}}-\lfloor\frac{m}{2}\rfloor-1\to\frac{1}{6}\\ \end{equation}

if this hold for all \(m\), it will hold for even \(m\), so we can replace \(m\) with \(2m\)

\begin{equation} \lim_{m\to\infty}\frac{4mb_{m}}{\pi b_{m}\sqrt{2m}-c_{m}}-m-1\to\frac{1}{6}\\ \end{equation}

It would appear that

\begin{equation} \frac{1}{2}\left(\frac{H_{\left\lfloor\frac{m}{2}\right\rfloor-\sqrt{-m-1}}-H_{\left\lfloor\frac{m}{2}\right\rfloor+\sqrt{-m-1}}}{\sqrt{-m-1}}+\frac{\pi\coth\left(\pi\sqrt{m+1}\right)}{\sqrt{m+1}}\right)=\frac{c_{m}}{a_{m}}\\ \end{equation}

then we have

\begin{equation} \lim_{m\to\infty}\frac{c_{m}}{a_{m}}=\lim_{m\to\infty}\frac{c_{m}}{2mb_{m}}=0\\ \end{equation}

\(\zeta(2)\)

we can do the same for \(\zeta(2)\) and we get a more complicated series of sums which equal expressions contatining cotangents of complex numbers.

Catalan’s Constant

Do

\begin{equation} 1+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)^{2}}=G\\ \end{equation}

Then the first sum plus the 1

\begin{equation} S=1+\frac{1}{8}\left(4-\sqrt{2}\pi\coth\left(\frac{\pi}{\sqrt{2}}\right)-\pi^{2}\csch\left(\frac{\pi}{\sqrt{2}}\right)^{2}\right)\\ \end{equation}

then

Primes

It is not nice, but we can also split the number line in terms of primes. We have that a sum over \(n\), is a sum over

\begin{equation} 1\\ \text{prime}(n)\\ \text{prime}(n+1)+1\\ \text{odd composites of form p+2}(n)\\ \text{odd composites of for p+2}(n)+1\\ \text{composite numbers not sum 2 primes}(n)\\ \text{composite numbers not sum 2 primes}(n)+1\\ \end{equation}

but apart from these, so far I think that covers all numbers! (I haven’t looked particularly far \(n\leq 5000\)).