Is series convergent? $$ \sum_{n=1}^{\infty } (-1)^{\frac{n(n+1)}{2}} \cdot \frac{1}{n}$$
Is it convergent? Of course it's not absolutely convergent, but I'm not sure about conditional convergence. Leibnitz seems not to work here.
 A: \begin{align}
\dfrac {n(n+1)}2&=\\
&n=4m+1&\rightarrow (4m+1)(2m+1)= odd\\
&n=4m+2&\rightarrow (2m+1)(4m+3)= odd\\
&n=4m+3&\rightarrow (4m+3)(2m+2)= even\\
&n=4m+0&\rightarrow 2m(4m+1)= even\\
\end{align}
\begin{align}
\therefore -\sum_{n=1}^{\infty } \frac{(-1)^{\frac{n(n+1)}{2}}}{n}&=\sum_{m=0}^{\infty} \left(\frac1{4m+1}-\frac1{4m+3}+\frac1{4m+2}-\frac1{4m+4}\right)\\
&=\sum_{m=0}^{\infty} \left(\frac{2}{(4m+1)(4m+3)}+\frac{2}{(4m+2)(4m+4)}\right)\\
&<\frac23+\frac14+\sum_{m=1}^{\infty} \left(\frac{2}{(4m)(4m)}+\frac{2}{(4m)(4m)}\right)
\\
&=\frac{11}{12}+\frac14\sum_{m=1}^{\infty}\frac1{m^2}=\frac{11}{12}+\frac14\cdot\frac{\pi^2}6\\
\end{align}

Dr. MV also mentioned that, this series can be expressed in a closed form as $\frac{\pi}4+\frac12\cdot\log 2$.
Then it came to me that we could do this way too (although dividing it into two series may be arguable):
$$\left(\frac11-\frac13+\frac15-\frac17+\frac19\cdots\right)+\left(\frac12-\frac14+\frac16-\frac18+\frac1{10}\cdots\right)$$ $$=\frac{\pi}4+\frac12\left(\frac11-\frac12+\frac13-\frac14+\frac15\cdots\right)=\frac{\pi}4+\frac12\cdot\log 2$$
A: I thought it might be instructive to see that the series can be evaluated in closed form.  The development is facilitated using the series representation of the Digamma Function 
$$\psi(x)=-\gamma+\sum_{n=0}^\infty\left(\frac{1}{n+1}-\frac{1}{n+x}\right)$$
where $\gamma=\lim_{N\to \infty}\left(\sum_{n=1}^N\frac1n -log(n)\right)$ is the Euler-Maschereoni Constant
We will also make use of Gauss's Digamma Theorem to determine the values of $\psi\left(\frac{\ell}{4}\right)$ for $\ell =1,2,3$.
Now, starting with the expression developed by @KayK. we have
$$\begin{align}
-\sum_{n=1}^\infty\frac{(-1)^{n(n+1)/2}}{n}&=\sum_{n=0}^\infty\left(\frac{1}{4n+1}+\frac{1}{4n+2}-\frac{1}{4n+3}-\frac{1}{4n+4}\right)\\\\
&=\frac14\sum_{n=0}^\infty \left(\frac{1}{n+1/4}-\frac{1}{n+1}\right)\\\\
&+\frac14\sum_{n=0}^\infty \left(\frac{1}{n+1/2}-\frac{1}{n+1}\right)\\\\
&-\frac14\sum_{n=0}^\infty \left(\frac{1}{n+3/4}-\frac{1}{n+1}\right)\\\\
&=\frac14\left(-\psi(1/4)-\psi(1/2)+\psi(3/4)+\psi(1)\right)\\\\
&=\frac14\left(\frac{\pi}{2}+3\log(2)+\gamma\right)\\\\
&+\frac14\left(2\log(2)+\gamma\right)\\\\
&+\frac14\left(\frac{\pi}{2}-3\log(2)-\gamma\right)\\\\
&+\frac14(-\gamma)\\\\
&=\frac{\pi}{4}+\frac12 \log(2)
\end{align}$$
