series convergence by decomposing it into smaller series I have to determine if the series $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} +1}{2}. (\frac{1+2i}{5})^{n} +\frac{(-1)^{n+1}+1}{2}.(\frac{2}{3})^{n} $$    is convergent or not.    This is what I tried:
I decomposed the above expression in 4 parts and wrote the series as:   $$ \sum_{n=1}^{\infty}{a}_{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{2}.(\frac{1+2i}{5})^{n} + \sum_{n=1}^{\infty} \frac{1}{2}.(\frac{1+2i}{5})^{n} +  \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2}.(\frac{2}{3})^{n}  + \sum_{n=1}^{\infty}\frac{1}{2} (\frac{2}{3})^{n} $$
and then determined the convergence of each of these four series by regular convergence tests- root test here, and all of them were found to be convergent, and then I concluded that their sum series must also be convergent.
So, my question: Is my  method right? and if it is, can somebody please suggest some other , better method of solving this problem and in case it is notcan somebody point out what is wrong?
Thanks in advance.
 A: Your series is absolutely convergent, thus convergent, since its sum $S$ is such that:
$$
\begin{align}
|S|&\leq  \sum_{n=1}^{\infty} \left(\left|\frac{(-1)^{n} +1}{2}\right| \left|\frac{1+2i}{5}\right|^{n} +\left|\frac{(-1)^{n+1} +1}{2}\right|\left|\frac{2}{3}\right|^{n}\right)\\\\
&\leq  \sum_{n=1}^{\infty} \left(\left|\frac{2}{2}\right| \left|\frac{\sqrt{5}}{5}\right|^{n} +\left|\frac{2}{2}\right|\left|\frac{2}{3}\right|^{n}\right)\\\\
&\leq  \sum_{n=1}^{\infty} \left( \left|\frac1{\sqrt{5}}\right|^{n} +\left|\frac{2}{3}\right|^{n}\right)<+\infty, \quad \left(\left|\frac1{\sqrt{5}}\right|<1,\,\left|\frac{2}{3}\right|<1\right).
\end{align}
$$
A: $$\sum_{n=1}^\infty \frac{(-1)^{n}}{2}.(\frac{1+2i}5)^n=\frac12\sum_{n=1}^\infty (\frac{-1-2i}5)^n=\frac12.\frac{-\frac{1+2i}5}{1+\frac{1+2i}{5}}$$   $$\sum_{n=1}^\infty \frac{1}{2}.(\frac{1+2i}5)^n=\frac12.\frac{\frac{1+2i}5}{1-\frac{1+2i}{5}}$$ $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2}.(\frac23)^n=-\frac12\sum_{n=1}^\infty (\frac{-2}3)^n=-\frac12 .\frac{-\frac23}{1+\frac23}$$ 
$$\sum_{n=1}^\infty \frac12(\frac{2}3)^n=\frac12.\frac{\frac23}{1-\frac23}$$
