show that the given series converges Prove that the series:
$$
\sum _{n=1} ^\infty \frac{(-1)^{\lfloor n/3 \rfloor}}{n}
$$
I tried to use Leibniz rule but I did not succeed. Any suggestions? Thanks for helping!

\begin{align}
\sum _{n=1}^{\infty} \frac{(-1)^{\lfloor n/3\rfloor}}{n} &= \frac{3}{2}+\sum _{n=1}^{\infty/3} \left[ \frac{(-1)^{n}}{3n} + \frac{(-1)^{n}}{3n+1} +\frac{(-1)^{n}}{3n+2} \right]\\
&=\frac{3}{2}+\sum _{n=1}^{\infty} (-1)^n \left(\frac{1}{3 n}+\frac{1}{3 n+1}  + \frac{1}{3 n+2}\right)\\
\end{align}
By the Alternating series test, the three series converge, see Paul's for further details. Below, we deduce their summation, using integration techniques.
\begin{align}
\hphantom{\sum _{n=1}^{\infty} \frac{(-1)^{\lfloor n/3\rfloor}}{n}} & \\&=
\frac{3}{2}+\left[-\frac{\log (8)}{9} \right] +
\left[ \frac{\pi \sqrt{3} + \log (8)}{9} - 1 \right] +
\left[ \frac{\pi \sqrt{3} - \log (8)}{9} - \frac{1}{2}\right] \\
&= \frac{2\pi \sqrt{3} - \log (8)}{9}.
\end{align}
