Welcome to our community

Be a part of something great, join today!

[SOLVED] n-epsilon proof

dwsmith

Well-known member
Feb 1, 2012
1,673
$\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$ if $a > 1$.

Not sure how to handle this one. Do I want have $\frac{1}{\sqrt[n]{\epsilon}} < a$?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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*}
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
$\lim\limits_{n\to\infty}\frac{1}{a^n} = 0$ 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
 

OhMyMarkov

Member
Mar 5, 2012
83
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.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Nice solution! It is by far more elementary than mine. :D Doesn't require the use of functions as the logarithm.