Can I use Limit Comparison Test to check series, $\sum\limits_{n=1}^{\infty}\frac{4^n}{3^{n-1}}$, for convergence? I’ve been trying to solve this problem and other similar problems.
I know that with LCT, if $b_n$ is divergent, (which $4^n$ is), then $\displaystyle\frac{a_n}{b_n}$ must be equal to infinity. However, evaluating the given gives me $0$ instead. 
What other tests can I use for this? Thank you!
 A: I see what you intended to do with limit comparison but the "intended" approach would be to rewrite this fraction in the following manner (I will only be looking at the fraction for the moment).
I begin by expanding the denominator and simplifying accordingly:
$$\frac{4^n}{3^{n-1}}=\frac{4^n}{(3^n)(3^{-1})}=\frac{3(4^n)}{3^n}$$
Next I will note the following:
$$\frac{3(4^n)}{3^n}=3\left(\frac{4}{3}\right)^n$$
I can now rewrite my series as follows:
$$\sum_{n=1}^{\infty} \frac{4^n}{3^{n-1}}=\sum_{n=1}^{\infty} 3\left(\frac{4}{3}\right)^n=3\sum_{n=1}^{\infty} \left(\frac{4}{3}\right)^n$$
Now it is a lot easier to see that we have a geometric series with $\frac{4}{3}$ being our $r$ or the common ratio. The convergence/divergence rules for geometric series are as follows:
(i) Converges with sum $s=\frac{a}{1-r}$ if $\vert r\vert\lt1$
(ii) Diverges if $\vert r\vert\geq1$
Thus since the absolute value of $r$ is indeed greater than $1$ the series $\sum_{n=1}^{\infty} \frac{4^n}{3^{n-1}}$ is divergent by geometric series test. QED (lol)
A: Observe that the $n^{th}$ term $a_n=3\frac{4^n}{3^n} > 1$ hence the series diverges. Because a necessary condition for series to converge is $n^{th}$ term should converge to zero.
