Unexpected result while calculation geometric series sum

Question

I have a geometric series like this :

$$\sum_{n=-\infty}^{-1}a^{-n}e^{-jwn}$$

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.

What is known about $a$ and $w$? I suppose $j=\mathrm e^{\tfrac{2\mathrm i\pi}3}$? — Bernard, May 17 '15 at 11:19
The formula you need is $\sum_{n=1}^\infty r^n=\frac{r}{1-r}$ — Gregory Grant, May 17 '15 at 11:26
@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. — jason, May 17 '15 at 11:27
It works with your formula, you need to plug in $m=1$ and then take the limit $b\rightarrow\infty$. — Gregory Grant, May 17 '15 at 11:33
@Jason: the imaginary root of $-1$ is denoted by $\mathrm i$, except by physicists (because is used to denote electric current). — Bernard, May 17 '15 at 11:33

score 2 · Accepted Answer · answered May 17 '15 at 11:39

2

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

answered May 17 '15 at 11:39

Bernard

175,478

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

Unexpected result while calculation geometric series sum

1 Answers1