Exact value of $\sum\limits_{n=1}^\infty(-1)^{n(n+1)/2}/n$?

Question

Wolfram is not computing it properly. What is the exact value of $$\sum_{n=1}^\infty\frac{(-1)^{n(n+1)/2}}{n}?$$ How to avoid imaginary $i$ coming from the exponent?

Answer 1

The series looks like

$$-\frac11-\frac12+\frac13+\frac14-\frac15-\frac16+\frac17+\frac18-\cdots$$

which may be reduced to

$$-\left (1-\frac13+\frac15-\cdots \right ) - \frac12 \left (1-\frac12+\frac13-\cdots \right )$$

which reduces to

$$-\frac{\pi}{4} - \frac12 \log{2}$$

Answer 2

Hint: $(-1)^{n(n+1)/2}$ will take two times the value $-1$, two times the value $1$ and so on.

Writing this as two sums should return $\quad-\dfrac{\log 2}2-\dfrac{\pi}4$.

Answer 3

Powers of $-1$ work modulo $2$, so you need to compute $$n(n+1)/2\mod 2$$ This thing has a period $4$: $$n=\{1,1,0,0,1,1,0,0,1,1,0,0,\ldots\}\mod 2$$

You can split into two sums and evaluate them separately.

Answer 4

Hint: Split the sum into four parts $$ \sum_{n=1}^\infty\frac{(-1)^{\frac{n(n+1)}{2}}}{n}= \sum_{n=0}^\infty\Big(-\frac{1}{4n+1} -\frac{1}{4n+2} +\frac{1}{4n+3} +\frac{1}{4n+4}\Big) $$ Then combine two parts with different signs. The partial sums up to $m$ can be expressed with the digamma function $\psi$ and the asymptotic series is $$-\frac{1}{2}\ln(2)-\frac{1}{4}\pi+ \frac{1}{4m}-\frac{9}{32m^2}+O(m^{-3})$$