Vtyjkyuk

Specific Question

Let $R$ be a whole number. Is it possible for any whole number $c>0$ that there is a nicer form for $sum_r=1^R^csum_r(-1)^r+d d^3$? 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:
$$f(x)=sum_r=1^xsum_r(-1)^r+d d^3$$
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 $fracpi^4384$ as $x$ gets large.

Exposition.

Let $phi_n(r)$ denote the number of integer solutions to the $n$-dimensional hypersphere $r=sum_i=1^n x_i^2$.

Then discovering $phi_2(r)=4 sum_r sin(fracpi2d)$ 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:

$$pi R^2 approxsum_r=1^R^2phi_2(r)$$ and after dividing both sides by $4R^2$ and unpacking our definitions we arrive $$fracpi4=lim_Rtoinfty frac1R^2sum_r=1^R^2sum_rsin Big(fracpi2d Big)$$
$$=sum_n=1^inftyfrac1nsinBig(fracpi2d Big)$$

So now that we've seen the development for a $2$-dimensional sphere what's the development for 8-dimensional sphere? Note that the $phi_8(r)= 16sum_r (-1)^r+dd^3$ which gives a way of approximate the volume of an $8$-dimensional hypersphere:

$$frac124pi^4 R^8 approx sum_r=1^R^2phi_8(r)$$

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

$$fracpi^4384= lim_Rtoinfty frac1R^8sum_r=1^R^2 sum_r (-1)^r+dd^3$$

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

$$fracpi^4384= lim_Rtoinfty frac1R^4csum_r=1^R^c sum_r (-1)^r+dd^3$$

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

$$1,frac816,frac3681,frac107256,frac233625,dots $$

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 $c=2$,

$$1,frac107256,frac21136561,frac1912865536, dots $$

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

$$sum_r=1^R^csum_r(-1)^r+d d^3$$

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 $n$?

Neat Question! What follows is mostly a long comment.

Define $tildesigma_s(n):=sum_d(-1)^d-1d^s$, an "alternating sum of divisors function." Your sum becomes $sum_r=1^x(-1)^r+1tildesigma_3(r)$.
If you now take the classical sum-of-divisors $sigma_s(n)$, then it's easy to see that $tildesigma_s(2k+1)=sigma_s(2k+1)$ and $tildesigma_s(2k)=sigma_s(2k)-2^s+1sigma_s(k)$.

Your sum is almost the famous Ramanujan Eisenstein sum $Q(q):=1+240sum_r=1^inftysigma_3(r)q^r=1+240sum_r=1^infty fracr^3q^r1-q^r$, 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 $tildesigma_3(n)$, along with analogous infinite Ramanujan Eisenstein sums:

Convolution Sums of some functions on divisors -- Hahn