Prove $(y_n)$ Converges to a Real Number Given $|y_{n+1}-y_n| \leq 2^{-n}$

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In summary: Why isn't it right that $$ 2^{-(m-1)}+ \dots+ 2^{-n} \leq (m-n) 2^{-n} \leq (n-(m-1)+1) 2^{-n}$$? (Thinking)
  • #1
evinda
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Hello! (Wave)

Let $(y_n)$ be a sequence of numbers such that $|y_{n+1}-y_n| \leq 2^{-n}$ for each $n \in \mathbb{N}$.
Show that the sequence $(y_n)$ converges to a real number.

Doesn't $|y_{n+1}-y_n| \leq 2^{-n}$ for each $n \in \mathbb{N}$ imply that $(y_n)$ is a Cauchy sequence?

So does it remain to show that every Cauchy sequence $(y_n)$ converges to a real number? If so how can we show this? (Thinking)
 
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  • #2
evinda said:
Hello! (Wave)

Let $(y_n)$ be a sequence of numbers such that $|y_{n+1}-y_n| \leq 2^{-n}$ for each $n \in \mathbb{N}$.
Show that the sequence $(y_n)$ converges to a real number.

Doesn't $|y_{n+1}-y_n| \leq 2^{-n}$ for each $n \in \mathbb{N}$ imply that $(y_n)$ is a Cauchy sequence?

So does it remain to show that every Cauchy sequence $(y_n)$ converges to a real number? If so how can we show this? (Thinking)

Hey evinda! (Smile)

If it's a Cauchy sequence, there's nothing more to do - it will converge to some real number.
However... it's not implied yet that it's a Cauchy sequence. (Worried)
 
  • #3
I like Serena said:
Hey evinda! (Smile)

If it's a Cauchy sequence, there's nothing more to do - it will converge to some real number.
However... it's not implied yet that it's a Cauchy sequence. (Worried)

A ok... I have thought the following:

We fix a $n \in \mathbb{N}$. We choose a $m \geq n+1$. Then

$|y_m-y_n|=|y_m-y_{m-1}+y_{m-1}+ \dots+ y_{n+1}-y_n| \overset{\text{ Triangle inequality}}{\leq} |y_m-y_{m-1}|+ \dots+ |y_{n+1}-y_n| \leq 2^{-(m-1)}+ \dots+ 2^{-n} \leq (n-m+2) 2^{-n}\leq 2^{-n+1}$.

This holds for any $n \in \mathbb{N}$ so the sequence is Cauchy.
Am I right? (Thinking)
 
  • #4
evinda said:
$(n-m+2) 2^{-n}\leq 2^{-n+1}$.

I think this doesn't hold. (Worried)
 
  • #5
I like Serena said:
I think this doesn't hold. (Worried)

Oh yes, right. It holds that $n \leq m-1$ and so $(n-m+2)2^{-n} \leq 2^{-n}$. Right?
 
  • #6
evinda said:
$|y_m-y_{m-1}|+ \dots+ |y_{n+1}-y_n| \leq 2^{-(m-1)}+ \dots+ 2^{-n} \leq (n-m+2) 2^{-n}\leq 2^{-n+1}$.

This holds for any $n \in \mathbb{N}$ so the sequence is Cauchy.
Am I right? (Thinking)

evinda said:
Oh yes, right. It holds that $n \leq m-1$ and so $(n-m+2)2^{-n} \leq 2^{-n}$. Right?

Hold on! (Wait)
Shouldn't it be:
$$|y_m-y_{m-1}|+ \dots+ |y_{n+1}-y_n| \leq 2^{-(m-1)}+ \dots+ 2^{-n} \leq (m-n) 2^{-n}$$
? (Wondering)
 
  • #7
I like Serena said:
Hold on! (Wait)
Shouldn't it be:
$$|y_m-y_{m-1}|+ \dots+ |y_{n+1}-y_n| \leq 2^{-(m-1)}+ \dots+ 2^{-n} \leq (m-n) 2^{-n}$$
? (Wondering)
Why isn't it right that

$$ 2^{-(m-1)}+ \dots+ 2^{-n} \leq (m-n) 2^{-n} \leq (n-(m-1)+1) 2^{-n}$$

? (Thinking)
 

FAQ: Prove $(y_n)$ Converges to a Real Number Given $|y_{n+1}-y_n| \leq 2^{-n}$

What is the definition of convergence in a sequence of real numbers?

The definition of convergence in a sequence of real numbers is that the sequence approaches a specific value, called the limit, as the index of the sequence gets larger and larger. In other words, the terms of the sequence get closer and closer to a single value as the sequence progresses.

How do we prove that a sequence of real numbers converges?

To prove that a sequence of real numbers converges, we must show that for any arbitrarily small positive value, there exists a point in the sequence after which all the values are within that small distance from the limit. This is known as the epsilon-delta definition of convergence.

What does the given condition $|y_{n+1}-y_n| \leq 2^{-n}$ tell us about the sequence?

This condition tells us that the difference between consecutive terms in the sequence is getting smaller and smaller as the index of the sequence increases. In other words, the sequence is becoming increasingly closer to a specific value as the index increases.

How does the given condition help us prove that the sequence $(y_n)$ converges?

The given condition helps us prove that the sequence $(y_n)$ converges by showing that the sequence is Cauchy. This means that for any arbitrarily small positive value, there exists a point in the sequence after which all the values are within that small distance from each other, indicating convergence towards a single limit.

Can we determine the limit of the sequence from the given condition?

No, we cannot determine the limit of the sequence solely from the given condition. We also need to know the starting value of the sequence in order to calculate the limit. However, the given condition does provide important information about the behavior of the sequence and helps us prove its convergence.

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