Proof That $\lim\limits_{n\to\infty}\frac{1}{a^n}=0$ if $a>1$

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In summary, to prove the limit $\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$, where $a > 1$, we need to show that there exists $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$, the expression $\left| \frac{1}{a^n} - 0 \right| < \varepsilon$ holds for all $\varepsilon > 0$. This can be done by taking $N_0 = \left\lceil \log_a \left( \frac{1}{\varepsilon} \right) \right\rceil$, and showing that $
  • #1
Dustinsfl
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$\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$ if $a > 1$.

Not sure how to handle this one. Do I want have $\frac{1}{\sqrt[n]{\epsilon}} < a$?
 
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  • #2
To prove this limit, we need to show that exists $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have that

$$\left| \frac{1}{a^n} - 0 \right| < \varepsilon,$$

for all $\varepsilon >0$.

Not sure what tools you have available, but if perhaps you could do

$$\frac{1}{a^n} < \varepsilon \leadsto a^n > \frac{1}{\varepsilon} \leadsto n \log_a a = n > \log_a \left( \frac{1}{\varepsilon} \right).$$

Therefore, take $N_0 = \left\lceil \log_a \left( \frac{1}{\varepsilon} \right) \right\rceil$.

Not entirely sure, but the whole process looks okay.
 
  • #3
Fantini said:
To prove this limit, we need to show that exists $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have that

$$\left| \frac{1}{a^n} - 0 \right| < \varepsilon,$$

for all $\varepsilon >0$.

Not sure what tools you have available, but if perhaps you could do

$$\frac{1}{a^n} < \varepsilon \leadsto a^n > \frac{1}{\varepsilon} \leadsto n \log_a a = n > \log_a \left( \frac{1}{\varepsilon} \right).$$

Therefore, take $N_0 = \left\lceil \log_a \left( \frac{1}{\varepsilon} \right) \right\rceil$.

Not entirely sure, but the whole process looks okay.

How do I now show $a > 1$?

Let $\epsilon > 0$ be given. Then $a^n < \frac{1}{\epsilon}$. Let's take the $\log_a$ of both sides.
Then
\begin{alignat}{1}
n\log_a a = n < \log_a\frac{1}{\epsilon}.
\end{alignat}
Let $N\in\mathbb{Z}$ such that $\log_a\frac{1}{\epsilon} < N$. For all $n > N$, we have that $\log_a\frac{1}{\epsilon} < N < n$.
\begin{alignat*}{3}
\log_a\frac{1}{\epsilon} & < & n\\
\frac{1}{\epsilon} & < & a^n\\
\left|\frac{1}{a^n} - 0\right| & < & \epsilon
\end{alignat*}
 
  • #4
dwsmith said:
$\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$ if $a > 1$.

Not sure how to handle this one. Do I want have $\frac{1}{\sqrt[n]{\epsilon}} < a$?
Since you said "if $a>1$", it seems like you're given this information. It is your hypothesis. It is because of this that we can take $\log_a r$. :D
 
  • #5
if $a>1$, we can write $a=1+y$ where $y>0$. We have $a^n = (1+y)^n > 1+ny>ny$ by the binomial law. Then, $\displaystyle \frac{1}{(1+y)^n}<\frac{1}{ny}$.

Claim: $\displaystyle \frac{1}{ny}$ goes to zero. Fix $\epsilon>0$, for $\forall n > \displaystyle \frac{1}{y\epsilon}$, we have $\displaystyle -\epsilon<0< \frac{1}{yn}<\epsilon$.

We have $\displaystyle 0<\frac{1}{a^n}<\frac{1}{ny}$, so $1/a^n$ converges to $0$ by the Sandwich theorem.
 
  • #6
Nice solution! It is by far more elementary than mine. :D Doesn't require the use of functions as the logarithm.
 

FAQ: Proof That $\lim\limits_{n\to\infty}\frac{1}{a^n}=0$ if $a>1$

What is a limit?

A limit, in mathematical terms, refers to a value that a function or sequence approaches as its input or index approaches a particular value or infinity. It is an important concept in calculus and is used to describe the behavior of functions and sequences at certain points or as they approach infinity.

How is the limit of a function or sequence determined?

The limit of a function or sequence is determined by evaluating the function or sequence at various points or indices close to the value or infinity it is approaching. This can be done analytically or graphically, and the resulting value is the limit of the function or sequence.

What does it mean for a limit to equal zero?

When a limit equals zero, it means that the function or sequence approaches zero as its input or index approaches a particular value or infinity. This can also be interpreted as the function or sequence becoming arbitrarily small or approaching zero as its input or index increases without bound.

How does the proof for $\lim\limits_{n\to\infty}\frac{1}{a^n}=0$ if $a>1$ work?

The proof for this limit involves using the definition of a limit and manipulating the expression $\frac{1}{a^n}$ to show that it approaches zero as $n$ approaches infinity. This is done by showing that the expression becomes arbitrarily small as $n$ increases without bound, thus satisfying the definition of a limit.

Why is it important to understand this limit?

Understanding this limit is important because it helps us understand the behavior of exponential functions with a base greater than 1. It also has many real-world applications, such as in finance and population growth, where exponential functions are commonly used. Additionally, this limit is a fundamental concept in calculus and lays the foundation for more advanced topics such as derivatives and integrals.

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