A note about this question: The original question asked seems likely impossible so I am really asking if we can exploit the technique below into giving us a 'nice' form for $\pi^4$. By nice form I mean an explicitly defined series of rationals.

A little explanation of the the technique.

The technique I use below I reference quite a bit on this site and therefore am putting a nice explanatory picture. The technique is a type of "Diophantine Approximation" I suppose. I don't know if the technique goes by a special name.

We are going to count the number integer solutions along some Diophantine equation $P(x,y)=n$. If let $\phi_n$ denote the number of solutions in $\mathbb{Z}$ to $P(x,y)=n$ then we may argue (under some conditions) that $\sum_{n=1}^N \phi_n$ approximates the area of the interior of $P(x,y)=N$. Let me give a concrete example. Let $P(x,y)=|x|^3+|y|^3$.

enter image description here

The actual area in the interior of $|x|^3+|y|^3=72$ is given by $\frac{72\Gamma(1/3)^3}{4\sqrt{3}\pi} \approx 63.59899501$ and this is weakly approximated by $69$ which is the total number of integers which are on some solution of $|x|^3+|y|^3=n$ as $n$ varies from $0$ to $72$. Approximating this area as the number of integer solutions should improve as a we let $n$ grow large.

Specific Question

Let $R$ be a whole number. Is it possible for any whole number $c>0$ that there is a nicer form for $\sum_{r=1}^{R^c}\sum_{d|r}(-1)^{r+d} d^3$? I can hope that we can just evaluate this sum but in lieu of that maybe something cheaper computationally. Taking $c=1$ we can examine: $$f(x)=\sum_{r=1}^{x}\sum_{d|r}(-1)^{r+d} d^3$$ Here is a table of the first 100 values of this function $f$. This function must grow $O(x^4)$ for reasons which should be clear after reading the exposition. TLDR: $f(x)/x^4$ limits to $\frac{\pi^4}{384}$ as $x$ gets large.

Edit 1: After thinking about this specific question I realized it's somewhat an unreasonable ask... In someways this specific question is tantamount to asking the Gauss circle problem but in 8 dimensions and not 2. It's seems that this problem doesn't have an answer for $2$-d case and it's not likely any easier as we increase the dimension. What I am really after is if we can exploit this summation to arrive at an expression for $\pi^4$ which isn't a divisor sum the same way we can do this in the 2 dimensional case. Full apologies: This is really a different question than the original query but considering the original query is likely impossible...


Let $\phi_n(r)$ denote the number of integer solutions to the $n$-dimensional hypersphere $r=\sum_{i=1}^n x_i^2$.

The 2-dimensional case.

Then discovering $\phi_2(r)=4 \sum_{d|r} \sin(\frac{\pi}{2}d)$ allows us to find a formula for $\pi$. Namely, Leibniz's formula for $\pi$. Considering that the sum of number of integer solutions for each $r$ from $1$ to $R^2$ should approximate the volume of a sphere (which for this special case $n=2$ goes by the special name 'circle') with radius $R$ we arrive at:

$$\pi R^2 \approx\sum_{r=1}^{R^2}\phi_2(r)$$ and after dividing both sides by $4R^2$ and unpacking our definitions we arrive $$\frac{\pi}{4}=\lim_{R\to\infty} \frac{1}{R^2}\sum_{r=1}^{R^2}\sum_{d|r}\sin \Big(\frac{\pi}{2}d \Big)=\sum_{n=1}^\infty{\frac{1}{n}\sin\Big(\frac{\pi}{2}n \Big)}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+ \dots $$

The second equality is sometimes referred to as a dirichlet convolution (I think). And it's this step that I can't seem to be able to take in $8$-dimensional case.

I will skip the 4-dimensional case.

But indeed, examining the 4 dimensional case we can achieve $\zeta(2)=\pi^2/6$. This can be found in the last proof in R. Chapman's compilation of proofs of the Basel problem. MSE has it's own compilation of these by the way.

So now that we've seen the development for a $2$ and $4$ dimensional sphere:

What's the development for 8-dimensional sphere?

Note that the ${\phi_8(r)= 16\sum_{d|r} (-1)^{r+d}d^3}$ which gives a way of approximate the volume of an $8$-dimensional hypersphere:

$$\frac{1}{24}\pi^4 R^8 \approx \sum_{r=1}^{R^2}\phi_8(r)$$

We divide both sides by $16R^8$ to arrive at

$$\frac{\pi^4}{384}= \lim_{R\to\infty} \frac{1}{R^8}\sum_{r=1}^{R^2} \sum_{d|r} (-1)^{r+d}d^3$$

In fact! It looks to me like for any whole number $c$,

$$\frac{\pi^4}{384}= \lim_{R\to\infty} \frac{1}{R^{4c}}\sum_{r=1}^{R^c} \sum_{d|r} (-1)^{r+d}d^3$$

This does provide a nice sequence which converges to $\pi^4/384$. Letting $c=1$ this sequence's rate of convergence is pretty atrocious.

$$1,\frac{8}{16},\frac{36}{81},\frac{107}{256},\frac{233}{625},\dots $$

The printed numbers are all accurate to ... zero decimal places... it should converge nonetheless.

but taking $c$ larger we arrive at this number much faster. For $c=2$,

$$1,\frac{107}{256},\frac{2113}{6561},\frac{19128}{65536}, \dots $$

Only the last number printed is accurate to the first decimal of $\pi^4/384$.

Can I see anything else here? Is there a closed form for this

$$\sum_{r=1}^{R^c}\sum_{d|r}(-1)^{r+d} d^3$$

Sequences of interest include A008457, A138503. I think this link may be a good resource but (I am pretty sure) the "formula" therein should be labelled as a generating function A055414.

I suppose it's unlikely that the problem is easier $c>1$ but I figured I would include this thought just in case the problem is somehow solvable for some specific $c$.

A more general food for thought question Does this development get me similar looking things for other dimensions $n$?

  • $\begingroup$ If anyone wanted to chime in about appropriate tags that would help too. $\endgroup$
    – Mason
    Aug 11, 2018 at 0:55
  • $\begingroup$ Wouldn't plugging $\sqrt[2]{R^c}$ into your formula proves your assertion? In fact it seems like the sum converges to $\frac{\pi^4}{384}$ for all real $c\ge 1$ $\endgroup$
    – cortek
    Aug 11, 2018 at 4:13
  • $\begingroup$ Ross Millikan's answer (and the subsequent comments) to this question may be related. $\endgroup$
    – Mason
    Aug 11, 2018 at 4:37
  • $\begingroup$ @cortek. Yeah. I am not so worried about this assertion. You are referring to "It looks to me like for any whole number c?" Note that $g(x)=\sum^{x}_{n=1}{a_n} \implies g(f(x))=\sum^{f(x)}_{n=1}a_n$. But the implication doesn't go both ways. So the problem may be easier for some specific $c$... but I doubt it. $\endgroup$
    – Mason
    Aug 11, 2018 at 4:44
  • 1
    $\begingroup$ I added a bounty with a mistype: should read $\pi^4$ not $\pi/4$. Whoops. It doesn't look like bounty text is editable. More here. $\endgroup$
    – Mason
    Oct 21, 2018 at 15:31

2 Answers 2


Neat Question! What follows is mostly a long comment.

Define $\tilde{\sigma}_s(n):=\sum_{d|n}(-1)^{d-1}d^s$, an "alternating sum of divisors function." Your sum becomes $\sum_{r=1}^x(-1)^{r+1}\tilde{\sigma}_3(r)$. If you now take the classical sum-of-divisors $\sigma_s(n)$, then it's easy to see that $\tilde{\sigma}_s(2k+1)=\sigma_s(2k+1)$ and $\tilde{\sigma}_s(2k)=\sigma_s(2k)-2^{s+1}\sigma_s(k)$.

Your sum is almost the famous Ramanujan Eisenstein sum $Q(q):=1+240\sum_{r=1}^\infty\sigma_3(r)q^r=1+240\sum_{r=1}^\infty \frac{r^3q^r}{1-q^r}$, the difference being your sum is finite and with the caveat that this sum blows up for $|q|=1$. That's unfortunate because for $|q|<1$, there are a slew of identites that $Q$ satisfies, and I'm not sure they will carry over easily in the finite case. It looks like this might be relevant paper, where you'll find a compendium of identities for $\tilde{\sigma}_3(n)$, along with analogous infinite Ramanujan Eisenstein sums:

Convolution Sums of some functions on divisors -- Hahn

  • $\begingroup$ Thanks! Just a note: $E_4$ in the notation of this wiki $\endgroup$
    – Mason
    Aug 11, 2018 at 6:42

Yes. We can exploit this technique into giving us a form of $\pi^4$. One thing to do with this is get a representation of $\zeta(4)$.

We are interested in interpreting $\sum_{n=1}^x r_8(n)$. One way is to think of it as an approximation of the $8$-dimensional ball which has volume $\pi^4/24$ and the other is to exploit a formula for $r_8(n)$ and examining the asymptotic behavior of $\sum_{n=1}^x r_8(n)$. This is what happens in this proof.

We will need that $\sum_{n=1}^x \sigma_k(x) \approx \frac{\zeta(k+1)} {k+1}x^{k+1}$ to make this very precise we may say that $$\lim_{x\to\infty}{\frac{k+1}{x^{k+1}}\sum_{n=1}^x{\sigma_k(n)}}=\zeta(k+1)$$ I think that this can be seen with just a slight rewriting of this.

We have built up all the needed tools I believe:

From equation (12) of this paper we can write


Now equipped with $\sum_{n=1}^x\sigma_3(\frac{n}{a})\approx \frac{\zeta(4)}{4}(\frac{x}{a})^4$

We may write that $$\begin{align} &\frac{\pi^4}{24}x^4 \\ &\approx&\sum_{n=1}^x r_8(n) \\ &=&16\sum_{n=1}^x {\sigma_3(n)} &-32\sum_{n=1}^x {\sigma_3(n/2)} &+256\sum_{n=1}^x {\sigma_3(n/4)}\\ &\approx &16 \frac{\zeta(4)}{4}x^4 &-32\frac{\zeta(4)}{4}(\frac{x}{2})^4 &+256 \frac{\zeta(4)}{4}(\frac{x}{4})^4 \\ &=x^4\zeta(4)\frac{15}{4} \end{align}$$

So we have that as we let $x$ grow large

$$\frac{\pi^4}{24}x^4 \approx \frac{15}{4}\zeta(4)x^4$$

So then we can conclude that $\zeta(4)=\pi^4/90$


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