Xrfgtjtk

What is the value of $$S=1-frac12-frac13+frac14+frac15+frac16-cdots$$ where the sign alternates over the triangular numbers?

To start, it is easy to prove convergence. The sum of each set of reciprocals (e.g. ${1/2,1/3}$ and ${1/4,1/5,1/6}$) can be written as $$sum_{i=1}^nfrac1{frac{n(n-1)}2+i},quad n=1,2,cdots$$ and it can be shown that it is monotonically decreasing (and tending towards zero) since begin{align}sum_{i=1}^nfrac1{frac{n(n-1)}2+i}>sum_{i=1}^{n+1}frac1{frac{n(n+1)}2+i}&impliedby sum_{i=1}^nfrac1{frac{n(n-1)}2+i}-sum_{i=1}^nfrac1{frac{n(n+1)}2+i}>frac2{n^2+3n+2}\&impliedby sum_{i=1}^nfrac 1{(n^2-n+2i)(n^2+n+2i)}>frac1{2n(n+1)(n+2)}\&impliedby frac n{(n^2-n+2cdot1)(n^2+n+2cdot1)}>frac1{2n(n+1)(n+2)}\&impliedby (n-2)(n^2+n+2)<2n^2(n+2)\&impliedby n^3+5n^2+4>0end{align} Is there a closed-form expression for the value of $S$?

In other terms we want to evaluate

$$ sum_{ngeq 1}(-1)^{n+1}left(H_{n(n+1)/2}-H_{n(n-1)/2}right)=int_{0}^{1}sum_{ngeq 1}(-1)^{n+1}frac{x^{n(n+1)/2}-x^{n(n-1)/2}}{x-1},dx$$
where the theory of modular forms ensures
$$ sum_{ngeq 0} x^{n(n+1)/2} = prod_{ngeq 1}frac{(1-x^{2n})^2}{(1-x^n)}=prod_{ngeq 1}frac{1-x^{2n}}{1-x^{2n-1}} $$
but I do not see an easy way for introducing a $(-1)^n$ twist in the LHS. On the other hand, the Euler-Maclaurin summation formula ensures
$$ H_n = log n + gamma + frac{1}{2n} - sum_{mgeq 2}frac{B_m}{m n^m} $$
in the Poisson sense. Replacing $n$ with $n(npm 1)/2$,
$$ H_{frac{n(n+1)}{2}}-H_{frac{n(n-1)}{2}} = logleft(tfrac{n+1}{n-1}right)+tfrac{2}{(n-1)n(n+1)}-sum_{mgeq 2}tfrac{2^m B_m}{m n^m}left(tfrac{1}{(n-1)^m}-tfrac{1}{(n+1)^m}right) $$
then multiplying both sides by $(-1)^n$ and summing over $ngeq 2$:
$$ sum_{ngeq 2}(-1)^nleft(H_{frac{n(n+1)}{2}}-H_{frac{n(n-1)}{2}} right)=\=5left(log(2)-tfrac{1}{2}right)-sum_{mgeq 2}tfrac{2^m B_m}{m}sum_{ngeq 2}(-1)^nleft(tfrac{1}{n^m(n-1)^m}-tfrac{1}{n^m(n+1)^m}right)
\=5left(log(2)-tfrac{1}{2}right)-sum_{mgeq 2}frac{2^m B_m}{m}left[frac{1}{2^m}-2sum_{ngeq 2}frac{(-1)^m}{n^m(n+1)^m}right]$$
where the innermost series is a linear combination of $log(2),zeta(3),zeta(5),ldots$ by partial fraction decomposition. This allows a reasonable numerical approximation of the original series and a conversion into a double series involving $zeta(2a)zeta(2b+1)$. I am not sure we can do better than this, but I would be delighted to be proven wrong.

Playing a bit with functions, a nice approximation of $sum_{ngeq 0}(-1)^n x^{n(n+1)/2}$ over $[0,1]$ is given by $frac{1}{x+1}-x^2(1-x)^2$, so the value of the original series has to be close to $log(2)-frac{1}{6}$. A better approximation of the function is $frac{1}{x+1}-x^2(1-x)^2+frac{3}{4}x^4(1-x)left(frac{4}{5}-xright)$, leading to the following improved approximation for the series: $log(2)-frac{53}{300}$. A further refinement,
$$ g(x)=sum_{ngeq 0}(-1)^n x^{n(n+1)/2} approx frac{1+x+2x^2}{1+2x+5x^2}$$
leads to $color{red}{Sapproxfrac{pi+3log 2}{10}}$. It might be interesting to describe how I got this approximation.
$g(0)$ and $g'(0)$ are directly given by the Maclaurin series, while $lim_{xto 1^-}g(x)=frac{1}{2}$ and $lim_{xto 1^-}g'(x)=-frac{1}{8}$ can be found through $mathcal{L}(f(e^{-x}))(s)$.
$g(x)$ is convex and decreasing on $(0,1)$ and any approximation of the
$$ frac{1+ax+(1+a)x^2}{1+(1+a)x+(2+3a)x^2}$$
kind with $a$ in a suitable range matches such constraint and the values of $g$ and $g'$ at the endpoints of $(0,1)$. We still have the freedom to pick $a$ in such a way that the derived approximation is both simple and accurate enough - I just picked $a=1$.