Convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{T_n+1}}{n},$ where $T_n$ is the $n$th Triangular number Consider the series $$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$ This is clearly not absolutely convergent. On the other hand, obvious choice, alternating series does not work here. Seems like the partial sum sequence is bounded but it is not monotone.
How can we prove that this series converges? and, where does it converge to? 

$$1\color{red}{+\frac12}\color{blue}{-\frac13}-\frac14+\frac15\color{cyan}{+\frac16}\color{magenta}{-\frac17}+\cdots$$
$$=1\color{red}{-\frac12+2\cdot\frac12}\color{blue}{+\frac13-2\cdot\frac13}-\frac14+\frac15\color{cyan}{-\frac16+2\cdot\frac16}\color{magenta}{+\frac17-2\cdot\frac17}+\cdots$$
$$=1-\frac12+\frac13-\frac14+\frac15-\cdots+2\left(\frac12-\frac13+\frac16-\frac17+\cdots\right)$$
$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}+2\sum_{n=1}^\infty\frac{1}{4n-2}-\frac{1}{4n-1}$$
$$=\ln(2)+\frac12\sum_{n=1}^\infty\frac{1}{n-1/2}-\frac{1}{n-1/4}$$
$$=\ln(2)+\frac12\left(H_{-1/4}-H_{-1/2}\right)$$
$$=\ln(2)+\frac12\left(\frac{\pi}{2}-\ln(2)\right)$$
$$=\frac{\pi}{4}+\frac12\ln(2)$$
note that we used the series representation of the harmonic number 
$$H_a=\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{n+a}\Longrightarrow H_a-H_b=\sum_{n=1}^\infty\frac{1}{n+b}-\frac{1}{n+a}$$
and we also used the results $H_{-1/4}=\frac{\pi}{2}-3\ln(2)$ and $H_{-1/2}=-2\ln(2)$ which can be obtained from the integral representation of the harmonic number $H_a=\int_0^1\frac{1-x^a}{1-x}dx$.
