Why Does $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$?

[Vali]

Why $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$ ?
Because $3^{n}> 2^{n-1} $ ?

 

[HOI]

Not that alone. $$\frac{2^{n-1}}{3^n}= \frac{1}{2}\left(\frac{2}{3}\right)^n$$ as well as the fact that $$a^n$$ "dominates" $$n^b$$ as n goes to infinity. That is, for any a and b, larger than 1, the limit of $$\frac{n^b}{a^n}$$, as n goes to infinity, is 0.

 

[Vali]

Thank you for the response!
Yes, I know that log<power<exponential<factorial but it's not completely clear for me in this case.I have exponential function at denominator and numerator too.That $2^{n-1}$ is exponential but $3^{n}$ is also an exponential function but is higher than the first one.I'm a little bit confused..

 

[I like Serena]

That's why we simplify it as:
$$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}
=\lim_{n\rightarrow \infty }n^{2016}\cdot \frac{ 2^{n-1}}{3^{n}}
=\lim_{n\rightarrow \infty }n^{2016}\cdot \frac 2 2 \cdot \frac{ 2^{n-1}}{3^{n}}
=\lim_{n\rightarrow \infty }n^{2016}\cdot \frac 1 2 \cdot \frac{ 2^{n}}{3^{n}}
=\lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \left(\frac{ 2}{3}\right)^n
$$
Now we can use that domination order as Country Boy explained, can't we?

 

[Vali]

I understood the simplification but I don't understand the form.I mean, I know that $\frac{n^b}{a^n}$ tends to 0 but in my form I have $n^{b}*a^{n}$.
In $\frac{n^b}{a^n}$ which is $n^{b}$ and which is $a^{n}$ ?

 

[I like Serena]

I understood the simplification but I don't understand the form.I mean, I know that $\frac{n^b}{a^n}$ tends to 0 but in my form I have $n^{b}*a^{n}$.
In $\frac{n^b}{a^n}$ which is $n^{b}$ and which is $a^{n}$ ?

We can rewrite what we have as:
$$\lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \frac{2^n}{3^n}
= \lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \frac{1}{\frac{3^n}{2^n}}
= \lim_{n\rightarrow \infty }\frac 1 2 \cdot \frac{n^{2016}}{\left(\frac{3}{2}\right)^n}
$$
Can we tell now which is $n^{b}$ and which is $a^{n}$ ?

 

[Vali]

I finally understood!
Thanks a lot!