Limit of the geometric sequence I have the following homework question:
For what values of $r$ does  $\lim_{n\rightarrow \infty} r^n$ exist? What does it converge to?
Correct me if I am wrong, but if $r$ is greater than or equal to 1, or less than or equal to -1, it will converge to infinity. However, if it is outside of that range, it will converge to negative or positive infinity.
 A: We examine the following limit:
$$
\displaystyle \lim_{n\to \infty} r^n
$$
If $|r| < 1$ then the the sequence will converge to $0$: can you see why? Try setting, say, $r = 1/2$ and write out the first few terms of $r^{1/2}$...can you see how the terms become increasingly small?
You are correct that when $r> 1$, $r^n \to \infty$.
When $r \leq -1$, the series will not converge to any number nor to infinity. The limit, i.e., does not exist. Because the exponent will shift from even to odd or odd to even with every increase in $n$, the sequence will oscillate from positive to negative, never approaching any number nor infinity. Note that for $r=-1$, the terms of the sequence are alternately $-1$ and $1$. If $r\lt -1$, and $n$ is large and even, then $r^n$ is large positive, while if $n$ is large odd, then $r^n$ is large negative.
Can you add what will happen when $r = 1$? That should be pretty obvious.
Summarizing, we have:
$$\lim_{n\rightarrow \infty} r^n =\left\{ 
 \begin{array}{rl}
 \infty & r>1\\ \\
 1  & r=1 \\ \\
 0 & r \in (-1,1)\\ \\
 \text{does not exist} & r \leq -1 \\ \\
\end{array}\right. $$
A: At first for $r>1$: We can write $r$ as $r=1+x$ with $x>0$ so we know with bernoulli 
$$ r^n=(1+x)^n \geq 1+n\cdot x$$. 
For $r=0$ the limit is obvious $0$. Now let $r\in (0,1)$, we can write 
$r=\frac{1}{p}$ with $p\in (1,\infty)$, 
So 
$$r^n = \frac{1}{p^n} = \frac{1}{(1+p-1)^n} \leq  \frac{1}{1+n (p-1)} \leq  \frac{1}{p-1} \cdot \frac{1}{n}$$
For $r=1$ the sequence converges to $1$ as it is always $1$.
Now we have to check $r<0$ 
$$r^n = (-1)^n (-r)^n$$ 
Which is a alternating sequence, so it converges when $(-r)^n$ converges to zero.
So all in all we see:
$$\lim_{n\rightarrow \infty} r^n =\left\{ 
 \begin{array}{rl}
 \infty & r>1\\
 1  & r=1 \\
 0 & r \in (-1,1)\\
 \text{ does not exists} & r \leq -1 \\
\end{array}\right. $$
A: You are correct about $r > 1$.
When $r \leq -1$ that sequence does not converge to $\infty$.  The magnitude of the number keeps getting bigger and bigger, but the sign also keeps switching so there is no convergence.
The remaining two cases are $r = 1$ and $-1 < r < 1$.  Both of these converge but not to $\infty$ or $-\infty$.  Try picking an $r$ and looking at the first $5$ or so terms in the sequence.  What's your guess as to what it converges to?
A: If $r\le -1$, then 
$$\lim_{n\to\infty} r^n$$
does not exist. The sequence does not converge. Note that for $r=-1$, the terms of the sequence are alternately $-1$ and $1$. If $r\lt -1$, and $n$ is large and even, then $r^n$ is large positive, while if $n$ is large odd, then $r^n$ is large negative. 
If $-1\lt r\lt 1$, then the sequence converges to $0$.
If $r=1$, then the sequence (clearly) converges to $1$.
For $r\gt 1$, there are some differences in the terminology that people use. You would have to check how the behaviour of the sequence $(r^n)$ is described in your book or notes.
Everyone of course agrees about what happens to $r^n$ as $n$ gets very large: it blows up.
Some people would say that in this case the sequence diverges, or does not converge. Because there is no real number $a$ such that $\lim_{n\to\infty} r^n=a$, they would say that the limit does not exist.
Some people would say the sequence diverges to $\infty$, or to $+\infty$
Some people would say that it converges to $\infty$, that the limit exists and is $\infty$, or $+\infty$. 
