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.

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solution is as follows:

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

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    $\begingroup$ $\frac{1-z^{N+1}}{1-z}$ is (for $z\ne 1$) $\sum_{n=0}^N z^n$, whereas in the exercise you have $\sum_{n=1}^N z^n$. $\endgroup$ Commented Jul 29, 2022 at 7:56
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    $\begingroup$ 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. $\endgroup$ Commented Jul 29, 2022 at 8:14
  • $\begingroup$ @insipidintegrator thank you for the hint. I will consider it next time. $\endgroup$
    – alfonso
    Commented Jul 29, 2022 at 8:38

1 Answer 1


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 $1$.

  • $\begingroup$ It is a must here to mention that $|z|\lt 1$. $\endgroup$ Commented Jul 29, 2022 at 8:51
  • $\begingroup$ Oops... I should have done that. Thanks! $\endgroup$
    – axr
    Commented Jul 29, 2022 at 9:17

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