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

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Discussion Overview

The discussion centers on evaluating the limit $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}$$ and understanding the behavior of the terms involved as \( n \) approaches infinity. Participants explore the dominance of exponential functions over polynomial functions in this context.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that since \( 3^{n} > 2^{n-1} \), this contributes to the limit being zero.
  • Another participant explains that the expression \( \frac{2^{n-1}}{3^n} \) can be rewritten as \( \frac{1}{2}\left(\frac{2}{3}\right)^n \) and notes that exponential functions dominate polynomial functions as \( n \) approaches infinity.
  • A participant expresses confusion regarding the comparison of the exponential functions in the numerator and denominator, noting that both are exponential but with different bases.
  • Further simplification of the limit is presented, breaking it down into a form that highlights the dominance of the exponential decay of \( \left(\frac{3}{2}\right)^n \) over the polynomial growth of \( n^{2016} \).
  • Participants discuss the form of the limit and seek clarification on identifying which terms correspond to \( n^{b} \) and \( a^{n} \) in the context of limits involving exponential decay.
  • One participant concludes that they have understood the concepts discussed, indicating a resolution to their confusion.

Areas of Agreement / Disagreement

Participants generally agree on the idea that exponential functions dominate polynomial functions in limits as \( n \) approaches infinity. However, there is some confusion regarding the specific forms of the expressions and how to apply the concepts correctly.

Contextual Notes

Participants express uncertainty about the application of the limit properties and the identification of terms in the context of exponential and polynomial growth. There are unresolved questions regarding the specific forms of the expressions used in the limit evaluation.

Vali
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Why $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$ ?
Because $3^{n}> 2^{n-1} $ ?
 
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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.
 
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..
 
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?
 
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 $a^{n}$ ?
 
Vali said:
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 $a^{n}$ ?
 
I finally understood!
Thanks a lot!
 

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