How to prove $a_n\to 0$ using the sequence location theorem?

  • MHB
  • Thread starter alexmahone
  • Start date
In summary, for a sequence where the ratio of consecutive terms tends to a value less than 1, it can be proved that the sequence is decreasing for a large enough value of n and that it also tends to 0. This can be shown using both indirect and direct proofs, with the direct proof using the fact that the sequence is bounded and the indirect proof relying on the assumption that the sequence does not tend to 0.
  • #1
alexmahone
304
0
Assume $\frac{a_{n+1}}{a_n}\to L$, where $L<1$ and $a_n>0$. Prove that

(a) $\{a_n\}$ is decreasing for $n\gg 1$;

I've done this part.

(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)

Indirect proof: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).

$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.

How do I go about the direct proof (one that doesn't use "proof by contradiction")?
 
Last edited:
Physics news on Phys.org
  • #2
Alexmahone said:
How do I go about the direct proof (one that doesn't use "proof by contradiction")?

Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
 
  • #3
Fernando Revilla said:
Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.

Is this a proof by induction that $|a_n|<\epsilon$? If so, how do we prove the base case for $n=1$?
 
Last edited:
  • #4
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
 
  • #5
HallsofIvy said:
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.

I used a slightly different approach:

$L<\frac{L+1}{2}$

$\frac{a_{n+1}}{a_n}<\frac{L+1}{2}$ for $n\gg 1$ (Using the "sequence location theorem")

So, $a_{n+1}< \left(\frac{L+1}{2}\right)a_n$ for $n\ge N$

$a_{N+k}<\left(\frac{L+1}{2}\right)^k a_N$ for $k\ge 1$ (Can be proved using induction over $k$.)

Since $\frac{L+1}{2}<1$, $a_{N+k}\to 0$ as $k\to\infty$.

So, $a_n\to 0$.
 
Last edited:

FAQ: How to prove $a_n\to 0$ using the sequence location theorem?

What is the definition of a sequence that tends to 0?

A sequence a_n is said to tend to 0 if for any positive real number ε, there exists a positive integer N such that for all n ≥ N, |a_n| < ε.

How do you prove that a sequence tends to 0?

To prove that a sequence a_n tends to 0, you need to show that for any positive real number ε, there exists a positive integer N such that for all n ≥ N, |a_n| < ε. This can be done by using the definition of limit and using appropriate mathematical techniques such as the squeeze theorem or the Cauchy criterion.

Is it possible for a sequence to tend to 0 but not converge?

Yes, it is possible for a sequence to tend to 0 but not converge. For example, the sequence {(-1)^n/n} tends to 0 as n approaches infinity, but it does not converge since it oscillates between -1 and 1.

Can a sequence that tends to 0 have negative terms?

Yes, a sequence that tends to 0 can have negative terms. For example, the sequence {(-1)^n/n} tends to 0 as n approaches infinity, but it has both positive and negative terms.

What is the importance of proving that a sequence tends to 0?

Proving that a sequence tends to 0 is important in many areas of mathematics and science. It allows us to understand the behavior of a sequence as its index increases, and it is often used in the convergence of series and in the study of limits. It is also important in applications such as differential equations, where the behavior of a solution can be determined by the behavior of a sequence that tends to 0.

Back
Top