Why is Taylor series expansion for 1 / ( 1 − x ) valid only...

Well, we have that
$s_n=\sum _{k=0}^n{x}^k=\frac{1-x^{n+1}}{1-x}$
Now, we look at $n\to \mathrm{\infty }$
$\underset{n\to \mathrm{\infty }}{lim}\frac{1-x^{n+1}}{1-x}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n{x}^k$
Clearly, we're only concerned about $x^{n+1}$. If $|x|>1$, then $x^{n+1}$ grows large, and there is no limit. If $|x|<1$, then $x-1\ne 0$ and $x^{n+1}$ goes to 0 as $n\to \mathrm{\infty }$, so in that case, we can assert that
$\underset{n\to \mathrm{\infty }}{lim}\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}=\sum _{k=0}^{\mathrm{\infty }}{x}^k$
If x=1, x−1=0 and we find ourselves in trouble. However, we can say that
$\sum _{k=0}^n{1}^k=n$
in which case the sequence of partial sums has no limits. Finally, for x=−1, we have $(-1{)}^{n+1}$ which oscillates and has no limit.
Thus,
$\frac{1}{1-x}=\sum _{k=0}^{\mathrm{\infty }}{x}^k$
makes sense only for |x|<1.