Difference between Fourier transform of $\left(\frac{1}{3}\right)^{-k}$ with $k\le -1$ and $\left(\frac{1}{3}\right)^{k}$ with $k\ge 1$ When I want to calculate the discrete Fourier transform of $\frac{1}{3}^{-k}$ with $k\le-1$ I get that:
$$
\sum_{k=-{\infty}}^{-1}\left(\frac{1}{3}\right)^{-k}e^{-j\omega k} = \sum_{k'=1}^{\infty} \left(\frac{1}{3}e^{j\omega}\right)^{k'} = \frac{\frac{1}{3}e^{j\omega}}{1-\frac{1}{3}e^{j\omega}}
$$
But if I change the variable before finding the Fourier transform I have the function $\left(\frac{1}{3}\right)^k$ with $k\ge1$ (both functions are the same). The discrete Fourier transform of the new function is as follows:
$$
\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}e^{-j\omega k} = \sum_{k=1}^{\infty}\left(\frac{1}{3}e^{-j\omega}\right)^{k} = \frac{\frac{1}{3}e^{-j\omega}}{1-\frac{1}{3}e^{-j\omega}}
$$
Are these two the same?
As Atul has pointed out, your both sequences are not the same, but they are time-reversed versions of each other:

k = np.arange(-15, 15)

s1 = (1/3.)**(-k)*(k &lt;= -1);
s2 = (1/3.)**(+k)*(k &gt;= +1)


plt.stem(k, s1, 'r')
plt.stem(k, s2, 'g')




Accordingly, what you see in your calculations is the time-reversal property of the Discrete-Time Fourier Transform:

$$x(t)=y(-t) \Leftrightarrow X(f)=Y(-f)$$

I.e. the frequency changes its sign in your calculations, as is expected from the time-reversal property.