Does $0.8^n$ converge or diverge and does it do so by monotonically or by oscillation? Solving for the limit of the sequence $0.8^n$, I first take the log of this limit.
I get limit approaches infinity $n(\ln0.8)$.
This this limit is infinity and diverges. Is this reasoning correct? And also how do you know if it does so monotonically or by oscillation?
Thanks so much.
 A: For $0\le r<1$, any sequence $$a_n=r^n$$
is monotonically decreasing, nonnegative, and converges to $0$.  There is no need for logarithms here, as OP is for the special case $r=0.8$.
A: $(0.8)^n$ is a geometric progression with first term $1$ and common ratio $r = 0.8 < 1$. So the sequence is convergent and it converges to $0$. The sequence is monotonically decreasing as is evident.
A: Hint: When you take the logarithm in a limit, you must also take the exponential.  In this case:
$$
\lim_{n\rightarrow\infty}0.8^n=e^{\lim_{n\rightarrow\infty}n\ln(0.8)}=e^{-\infty}.
$$
Therefore, even though the exponent diverges, the value of the original limit does not diverge.
A: Write $0.8=\frac{4}{5}=\frac{1}{1+\frac{1}{4}}$ and analyse $\left(\frac{1}{1+\frac{1}{4}}\right)^n=\frac{1}{\left(1+\frac{1}{4}\right)^n}$.
Can you see that $\left(1+\frac{1}{4}\right)^n$ increases monotonically? 
If you can see that the answer to this question  is yes, then  you could see that $0.8^n=\frac{1}{\left(1+\frac{1}{4}\right)^n}$ is decreasing monotonically.
