DTFT question regarding a pair

[Pavoo]

Homework Statement

My book writes the following: using pair for the Discrete Time Fourier Transform:
-a^{k}u[-k-1] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | > 1

Homework Equations

Well, for the simple similar pair such as:

a^{k}u[k] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | < 1

The calculation is pretty straightforward regarding the GP series.
In the above however, I get lost where they get the result from.

The Attempt at a Solution

I've come to here, coming from DTFT definition and simplifying the unitstep boundaries to the sum of:
\sum_{1}^{\infty}a^{-k}e^{iwk}

The question is, how do I get from here to the above?

a^{k}u[k] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | < 1

I am thankful for any walk through you can give, since I spent way too much time on this problem. Is there something obvious I don't see about this one, or is the calculation of GP series I am mistaken about?

Thanks in advance!

 

[Pavoo]

Solved

Finally I understood where I've done wrong. For anyone's interest, here it is:

\sum_{1}^{\infty}a^{-k}e^{iw}=-\frac{a^{-1}e^{iw}}{1-a^{-1}e^{iw}}

From here one can solve the thing easily, which gives the correct condition that a>1 as well.

The thread may be locked now.

 

[HallsofIvy]

That is wrong unless you have written the sum on the left incorrectly.

That is, of course, using the formula for the sum of a geometric series:
a\sum_{n=0}^\infty r^n= \frac{a}{1- r}

\sum_{k=1}^\infty a^{-k}e^{ik\omega}= a^{-1}e^{ik\omega}\sum_{k=0}^\infty (a^{-1}e^{k\omega})^k= \frac{a^{-1}e^{i\omega}}{1- a^{-1}e^{i\omega}}

But without the "k" in the exponential, it would be just
\sum_{k= 1}^\infty a^{-k}e^{i\omega}= a^{-1}e^{i\omega}\sum_{k=0}^\infty a^{-k}= \frac{a^{-1} e^{i\omega}}{1- a^{-1}}