Find $\lim_{n\to\infty} 16^{(n/2) }/ 3^{2n+1}$ I have problems with the following task: Find $\lim_{n\to\infty} 16^{(n/2) }/ 3^{2n+1}$.
I would start with taking the log an then dividing by n, however, then  I would end up with a false solution:
$$\lim_{n\to\infty} 16^{(n/2) }/ 3^{2n+1}$$  $$\lim_{n\to\infty}   (n/2)\ln(2) - (2n+1)\ln(3)= $$ $$\lim_{n\to\infty}   (1/2)\ln(2) - (2)\ln(3) - (1/n)\ln(3)= $$ 
$$ (1/2)\ln(2) - (2)\ln(3) $$
Thanks for your help!
 A: Simplifying the expression,
$$\frac{16^{n/2}}{3^{2n+1}} = \frac{\left(16^{1/2}\right)^n}{3 \cdot \bigl(3^{2}\bigr)^n} = \frac{4^n}{3\cdot9^n} = \frac{1}{3}\left(\frac{4}{9}\right)^n$$
Therefore,
$$\lim_{n\to\infty}\frac{16^{n/2}}{3^{2n+1}} = \frac{1}{3} \lim_{n\to\infty}\left(\frac{4}{9}\right)^n = 0$$
since $0 < 4/9 < 1$. 

Using logs isn't as nice, but if you suppose the limit is $L$, you'll get to an expression similar to this:
$$\lim_{n\to\infty} n(\ln 4 - 2 \ln 3) = \lim_{n\to\infty} n \ln (4/9) = \ln (3L).$$
Since $4/9 < 1$, it's natural log is negative and hence the left side goes to $-\infty$. Since
$$\lim_{x \to 0^+} \ln (x) = -\infty$$
we can conclude $3L = 0$, which implies $L = 0$.
A: Via logarithm: 
$$\lim_{n \to \infty} \ln \left( \frac{16^{n/2}}{3^{2n+1}} \right) = \lim_{n \to \infty} \left( \frac{n}{2} \ln(16) - (2n+1) \ln(3) \right) 
=\lim_{n \to \infty} n \left(\frac{\ln(16)}{2}-2\ln(3)\right)-\ln(3)=-\infty$$
since $\frac{\ln(16)}{2}-2\ln(3)<0$ and therefore $$\lim_{n \to \infty} \frac{16^{n/2}}{3^{2n+1}} = 0$$
A: Even if there are simpler solutions (as shown in answers and comments), you were almost on the right track.
Let me do it using you way considering $$a_n= \frac n 2\ln(2) - (2n+1)\ln(3)$$ and, as you almost did, factor the $n$ to get $$a_n= n\left(\frac 1 2\ln(2) - (2+\frac 1n)\ln(3)\right)$$ When $n$ become very large $\frac 1n$ becomes negligible when comared to $2$. This means that $$a_n \sim n\left(\frac 1 2\ln(2) - 2\ln(3)\right)$$ and the quantity inside brackets is negative. So $$\lim_{n\to\infty}a_n=-\infty\implies\lim_{n\to\infty}e^{a_n}=0$$
A: $16^{n/2}/3^{2n+1}=4^n/9^n*3=\frac 13\frac  49 ^n $
So the limit is $0$
