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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 $$\sum_{i=1}^n\frac1{\frac{n(n-1)}2+i},\quad n=1,2,\cdots$$ and it can be shown that it is monotonically decreasing (and tending towards zero) since \begin{align}\sum_{i=1}^n\frac1{\frac{n(n-1)}2+i}>\sum_{i=1}^{n+1}\frac1{\frac{n(n+1)}2+i}&\impliedby \sum_{i=1}^n\frac1{\frac{n(n-1)}2+i}-\sum_{i=1}^n\frac1{\frac{n(n+1)}2+i}>\frac2{n^2+3n+2}\\&\impliedby \sum_{i=1}^n\frac 1{(n^2-n+2i)(n^2+n+2i)}>\frac1{2n(n+1)(n+2)}\\&\impliedby \frac n{(n^2-n+2\cdot1)(n^2+n+2\cdot1)}>\frac1{2n(n+1)(n+2)}\\&\impliedby (n-2)(n^2+n+2)<2n^2(n+2)\\&\impliedby n^3+5n^2+4>0\end{align} Is there a closed-form expression for the value of $S$?

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    $\begingroup$ If you're curious, the general term can be written as $$\frac{(-1)^{\lfloor \sqrt{2n-7/4}+1/2\rfloor}}{n}$$ $\endgroup$ – Frpzzd Dec 10 '18 at 20:52
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    $\begingroup$ If we define the function $$f(x)=1-x-x^2+x^3+x^4+x^5-...$$ Then we have that $$\frac{1-x}{2}f(x)=\sum_{n=1}^\infty (-1)^n x^{n(n+1)/2}$$ Which suggests some sort of representation in terms of [Theta Functions][1]. From here, we have that the sum you're asking about can be written as $$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...=\int_0^1 f(x)dx$$ Perhaps, after seeing this, someone who knows more about Theta functions can swoop in and simplify this to something more useful or satisfying. [1]: en.wikipedia.org/wiki/Theta_function $\endgroup$ – Frpzzd Dec 10 '18 at 21:06
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    $\begingroup$ @Frpzzd That sum is actually $[(1-x)/2] f(x) -1/2$. $\endgroup$ – eyeballfrog Dec 10 '18 at 21:50
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    $\begingroup$ @Frpzzd It seems it's really $[(1-x)/2]f(x) + 1/2$, and it's too late for me to edit my comment too. $\endgroup$ – eyeballfrog Dec 10 '18 at 22:22
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    $\begingroup$ Hey there, @TheSimplifire ! I am suspended from chat until later Tuesday, so I can not respond in the chatroom in which you pinged me (thought I can still get pings, and read everything.) Wow!! I'm sorry about the hammer on the area 51 math challenges proposal. I know you've put a heck of a lot of time into navigating the proposal, and site, to make it viable. But don't take it personally. And you've got the kind of motivation and determination to see possibly a different proposal through, or, to become a mod on MSE!! ;-) $\endgroup$ – Namaste Dec 16 '18 at 20:06
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In other terms we want to evaluate

$$ \sum_{n\geq 1}(-1)^{n+1}\left(H_{n(n+1)/2}-H_{n(n-1)/2}\right)=\int_{0}^{1}\sum_{n\geq 1}(-1)^{n+1}\frac{x^{n(n+1)/2}-x^{n(n-1)/2}}{x-1}\,dx$$ where the theory of modular forms ensures $$ \sum_{n\geq 0} x^{n(n+1)/2} = \prod_{n\geq 1}\frac{(1-x^{2n})^2}{(1-x^n)}=\prod_{n\geq 1}\frac{1-x^{2n}}{1-x^{2n-1}} $$ 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 $$ H_n = \log n + \gamma + \frac{1}{2n} - \sum_{m\geq 2}\frac{B_m}{m n^m} $$ in the Poisson sense. Replacing $n$ with $n(n\pm 1)/2$, $$ H_{\frac{n(n+1)}{2}}-H_{\frac{n(n-1)}{2}} = \log\left(\tfrac{n+1}{n-1}\right)+\tfrac{2}{(n-1)n(n+1)}-\sum_{m\geq 2}\tfrac{2^m B_m}{m n^m}\left(\tfrac{1}{(n-1)^m}-\tfrac{1}{(n+1)^m}\right) $$ then multiplying both sides by $(-1)^n$ and summing over $n\geq 2$: $$ \sum_{n\geq 2}(-1)^n\left(H_{\frac{n(n+1)}{2}}-H_{\frac{n(n-1)}{2}} \right)=\\=5\left(\log(2)-\tfrac{1}{2}\right)-\sum_{m\geq 2}\tfrac{2^m B_m}{m}\sum_{n\geq 2}(-1)^n\left(\tfrac{1}{n^m(n-1)^m}-\tfrac{1}{n^m(n+1)^m}\right) \\=5\left(\log(2)-\tfrac{1}{2}\right)-\sum_{m\geq 2}\frac{2^m B_m}{m}\left[\frac{1}{2^m}-2\sum_{n\geq 2}\frac{(-1)^m}{n^m(n+1)^m}\right]$$ 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 $\sum_{n\geq 0}(-1)^n x^{n(n+1)/2}$ 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 $\frac{1}{x+1}-x^2(1-x)^2+\frac{3}{4}x^4(1-x)\left(\frac{4}{5}-x\right)$, leading to the following improved approximation for the series: $\log(2)-\frac{53}{300}$. A further refinement, $$ g(x)=\sum_{n\geq 0}(-1)^n x^{n(n+1)/2} \approx \frac{1+x+2x^2}{1+2x+5x^2}$$ leads to $\color{red}{S\approx\frac{\pi+3\log 2}{10}}$. It might be interesting to describe how I got this approximation.
$g(0)$ and $g'(0)$ are directly given by the Maclaurin series, while $\lim_{x\to 1^-}g(x)=\frac{1}{2}$ and $\lim_{x\to 1^-}g'(x)=-\frac{1}{8}$ 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 $a=1$.

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