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

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

and so on

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

the first coefficients of the latte sums take the formula

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

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

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

so we can rewrite

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

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

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

The first term is easy

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

the rest is then

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

with $$\psi_{n}$$ the polygamma function, and

$$t_{+}=\frac{3}{2}+n+\frac{1}{2}\sqrt{1+8n}\\ t_{-}=\frac{3}{2}+n-\frac{1}{2}\sqrt{1+8n}\\$$

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

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

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

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

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

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

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

Sum

$$\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1\\$$

so then when we perform

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

for $$m=1$$ we get

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

Generating Functions

We may write the set of powers of $$x$$ as generating functions as follows

$$T=\frac{1}{2}\theta_{2}(0,\sqrt{x})\\$$

for elliptic theta, then

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

and the general term becomes

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

Thus

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

where

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

where $$A+1$$ is the beautiful series

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

with $$n$$ of each coefficient! This then gives

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

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

$$n^{2}\\ n^{2}+1\\ n^{2}+2\\ (n+1)^{2}+3\\ (n+1)^{2}+4\\ (n+2)^{2}+5\\ ..\\$$

Then

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

We can take the divergent sum

$$Q=\sum_{n=1}^{\infty}\frac{1}{n}\\$$

and then go through this process and state that

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

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

$$\frac{1}{\lfloor\frac{m}{2}\rfloor+1+\frac{1}{6}}\\$$

for some term

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

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

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

but we know that

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

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

$$\frac{a_{m}}{2m}=\{1,1,2,5,9,35,88\}=b_{m}\\$$

where $$b_{m}$$ appears more basic than $$a_{m}$$, so we can write

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

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

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

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

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

It would appear that

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

then we have

$$\lim_{m\to\infty}\frac{c_{m}}{a_{m}}=\lim_{m\to\infty}\frac{c_{m}}{2mb_{m}}=0\\$$

$$\zeta(2)$$

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

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

Then the first sum plus the 1

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

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

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

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