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 $S$?

  • 4
    $\begingroup$ If you're curious, the general term can be written as $$\frac{(-1)^{\lfloor \sqrt{2n-7/4}+1/2\rfloor}}{n}$$ $\endgroup$ – Franklin Pezzuti Dyer Dec 10 '18 at 20:52
  • 2
    $\begingroup$ 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 $\endgroup$ – Franklin Pezzuti Dyer Dec 10 '18 at 21:06
  • 1
    $\begingroup$ @Frpzzd That sum is actually $[(1-x)/2] f(x) -1/2$. $\endgroup$ – eyeballfrog Dec 10 '18 at 21:50
  • 3
    $\begingroup$ @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
  • 1
    $\begingroup$ 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!! ;-) $\endgroup$ – amWhy Dec 16 '18 at 20:06

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 $a=1$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.