Why is this a valid condition for convergence? We know that $\sum_{i=1}^nx^n=\frac{1-x^{n+1}}{1-x}$, for $n\neq 1$. Why are we allowed to say that as $n\rightarrow\infty$, the terms $x^n\rightarrow 0$ if and only if $|x|<1$, so the sum converges iff $|x|<1$? Obviously in order to converge the terms must go to zero but how is that a sufficient condition? For example, in the harmonic series the terms go to zero but the sum diverges.
 A: $$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$$
This means that the value of the right is exactly equal to that on the right and most importantly the right converges if and only if the left converges.
For any x, $\frac{1-x^{n+1}}{1-x} = \frac{1}{1-x} - \frac{x^{n+1}}{1-x}$ so $\frac{1-x^{n+1}}{1-x}$ converges if and only if both $\frac{1}{1-x}$ and $\frac{x^{n+1}}{1-x}$ converge. $\frac{1}{1-x}$ obviously converges for all x as it is independent of n so $\frac{1-x^{n+1}}{1-x}$ converges if and only if $\frac{x^{n+1}}{1-x}$ converges
$$ \lim_{n \rightarrow \infty} \frac{x^{n+1}}{1-x} = \frac{1}{1-x}\lim_{n \rightarrow \infty} x^{n+1}$$
So $\frac{x^{n+1}}{1-x}$ converges if and only if $x^{n+1}$ converges which converges for $|x| \lt 1$. So to answer the last question, it isn't a sufficient condition but in this specific case the condition for the terms to approach 0 happens to be the same as the condition for the closed form of the sum to converge
A: hint
$$x>0\implies x^n=e^{n\ln(x)}$$
So,
$$\lim_{n\to+\infty}x^n=0\iff \ln(x)<0$$
$$\iff x<1$$
and
$$x<0\implies x^n=(-1)^ne^{n\ln(-x)}$$
