This is something in the textbook I am reading:
Example 1.30. The geometric series $\sum_{1}^{\infty} x^{k}$ is convergent to $(1-x)^{-1}$ if $|x|<1$ and divergent otherwise.
My question is : isn't $\sum_{0}^{\infty} x^{k}$ is convergent to $(1-x)^{-1}$ if $|x|<1$ and $\sum_{1}^{\infty} x^{k}$ is convergent to $\frac{1}{1-x}-1=\frac{x}{1-x}$ if $|x|<1$ Thank you in advance for the help :)