I have a geometric series like this :


When I make $m = -n$ substituion, it becomes this : $$\sum_{m=1}^{\infty}(ae^{jw})^m$$

And when I calculate summation, the result becomes : $ae^{jw}/(1-ae{^jw})$

I can't understand why we end up with this result. I know that geometric series solution is this :

enter image description here

But when I apply this formula, I can't find this result. How do we calculate the nominator of the result as ae^jw ? a and w are constants and j is the imaginary number, namely square root of -1 Thanks.

  • $\begingroup$ What is known about $a$ and $w$? I suppose $j=\mathrm e^{\tfrac{2\mathrm i\pi}3}$? $\endgroup$ – Bernard May 17 '15 at 11:19
  • $\begingroup$ @Bernard updated my question. Thanks. $\endgroup$ – jason May 17 '15 at 11:21
  • $\begingroup$ The formula you need is $\sum_{n=1}^\infty r^n=\frac{r}{1-r}$ $\endgroup$ – Gregory Grant May 17 '15 at 11:26
  • $\begingroup$ @GregoryGrant what is wrong with the formula I have? It's a general one and should work with this too. Thanks for the edits btw. $\endgroup$ – jason May 17 '15 at 11:27
  • $\begingroup$ It works with your formula, you need to plug in $m=1$ and then take the limit $b\rightarrow\infty$. $\endgroup$ – Gregory Grant May 17 '15 at 11:33

Since $\lim\limits_{b\to+\infty}r^{b+1}=0$ if $\lvert r\rvert<1$, we have: $$\lim_{b\to\infty}\sum_{k=1}^br^k=\lim_{b\to\infty}\frac{r-r^{b+1}}{1-r}=\frac r{1-r}. $$

  • $\begingroup$ You have yet to show that $|ae^{jw}|<1$. $\endgroup$ – Demosthene May 17 '15 at 11:51
  • $\begingroup$ @Demosthens: I only posted a sketch to help first, not a complete solution. $\endgroup$ – Bernard May 17 '15 at 11:56

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