# How to solve this problem with geometrical series?

I have this question from coursera tutorial. Howver, it has been a long time that I did nothing with maths and cannot solve this problem although I found the formula of geometric series. It is important to understand it. Therefore, I will appreciate a step by step solution and help.

Let $$s[n]=\displaystyle \frac{1}{2^n}+j\frac{1}{3^n}$$.Compute $$\displaystyle\sum_{n=1}^{\infty}s[n]$$.

Formula is as below but I couldn't manage to solve with complex numbers.

solution is as follows:

but I couldn't understand how 1/2 and 1/3 comes.

• $\frac{1-z^{N+1}}{1-z}$ is (for $z\ne 1$) $\sum_{n=0}^N z^n$, whereas in the exercise you have $\sum_{n=1}^N z^n$. Commented Jul 29, 2022 at 7:56
• Hi! Since you are new here, I wanted to let you know that using pictures for critical portions of the post (except diagrams, of course) is discouraged. Please learn mathjax to write out the math. Commented Jul 29, 2022 at 8:14
• @insipidintegrator thank you for the hint. I will consider it next time. Commented Jul 29, 2022 at 8:38

The sum of first $$n$$ terms of a geometrical series is given by $$S_n = a \; \frac{1-z^{n}}{1-z}$$ where $$a$$ is the first term. For $$n \rightarrow \infty$$ $$S_\infty = \frac{a}{1-z}$$ ONLY for $$|z|<1$$. Since your problem wants us to start the summation from $$n=1$$, rather than $$n=0$$, the first terms will be $$1/2$$ and $$1/3$$ respectively rather than $$1$$ and $$1$$.
• It is a must here to mention that $|z|\lt 1$. Commented Jul 29, 2022 at 8:51