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

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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

I think this doesn't hold. (Worried)

 

[evinda]

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?

 

[I like Serena]

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

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

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)

 

[evinda]

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)