Lim sup proof

[fishturtle1]

Homework Statement

Prove the $\limsup \vert s_n \vert = 0$ iff $\lim s_n = 0$.

Homework Equations

$\limsup s_n = \lim_{N\rightarrow \infty} \sup \lbrace s_n : n > N \rbrace = \sup \text{S}$

$\liminf s_n = \lim_{N\rightarrow \infty} \inf \lbrace s_n : n > N \rbrace = \inf \text{S}$

Definition of limit: $\lim s_n = L$ iff For all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that $n > N(\varepsilon)$ implies $\vert s_n - L \vert < \varepsilon$.

The Attempt at a Solution

$(\leftarrow)$ Suppose $\lim s_n = 0$. Then for all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that $\vert s_n - 0 \vert = \vert s_n \vert < \varepsilon$. So $\vert \vert s_n \vert \vert < \varepsilon$. So $\lim \vert s_n \vert = 0$. Since $\lim \vert s_n \vert = 0 \epsilon \mathbb{R}$, we can say $\limsup \vert s_n \vert = 0$.

$(\rightarrow)$ Suppose $\limsup \vert s_n \vert = 0$.

 

[fresh_42]

Homework Statement

Prove the $\limsup \vert s_n \vert = 0$ iff $\lim s_n = 0$.

Homework Equations

$\limsup s_n = \lim_{N\rightarrow \infty} \sup \lbrace s_n : n > N \rbrace = \sup \text{S}$

$\liminf s_n = \lim_{N\rightarrow \infty} \inf \lbrace s_n : n > N \rbrace = \inf \text{S}$

Definition of limit: $\lim s_n = L$ iff For all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that $n > N(\varepsilon)$ implies $\vert s_n - L \vert < \varepsilon$.

The Attempt at a Solution

$(\leftarrow)$ Suppose $\lim s_n = 0$. Then for all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that $\vert s_n - 0 \vert = \vert s_n \vert < \varepsilon$. So $\vert \vert s_n \vert \vert < \varepsilon$. So $\lim \vert s_n \vert = 0$. Since $\lim \vert s_n \vert = 0 \epsilon \mathbb{R}$, we can say $\limsup \vert s_n \vert = 0$.
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$(\rightarrow)$ Suppose $\limsup \vert s_n \vert = 0$.

The step from $\lim |s_n|=0$ to $\sup S = 0$ is a bit of a cheat. You have $|s_n| < \varepsilon$ for all $n > N_\varepsilon$. Now can you choose a series of $\varepsilon$ such that $\{|s_n|\, : \,n> N_\varepsilon\}$ is bounded in a way that shows that these boundaries converge to zero?

This should also be your strategy in the other direction: use the definitions and make yourself clear, what has to be shown. Just claiming things isn't a proof, although the result is expected.

 

[fishturtle1]

The step from $\lim |s_n|=0$ to $\sup S = 0$ is a bit of a cheat. You have $|s_n| < \varepsilon$ for all $n > N_\varepsilon$. Now can you choose a series of $\varepsilon$ such that $\{|s_n|\, : \,n> N_\varepsilon\}$ is bounded in a way that shows that these boundaries converge to zero?

This should also be your strategy in the other direction: use the definitions and make yourself clear, what has to be shown. Just claiming things isn't a proof, although the result is expected.

Thanks for the reply,

Since $\lim s_n = 0$, the supremums of $(s_k)$ where $k \ge n$ must be getting smaller and smaller since $s_k$'s are getting smaller and smaller...

Thm. 11.4 from the text: Every sequence $(s_n)$ has a monotonic subsequence.

So consider a monotonic subsequence $s_{n_k}$ that converges to 0. Then $\limsup s_{n_k} = 0$. Is this what you meant? Edit: no. This is just the same argument I used before... So should I look for a subsequence of $N_\varepsilon$ or a subsequence of $s_n$?

 

[fresh_42]

I'm not sure, if I make it too complicated or not. Say $S_N := \{s_n\, : \,n>N\}$ and $\varepsilon >0$. Then there is a $N$ with $|s_n|<\varepsilon$ for all $n>N$, if we assume $\lim_{n \to \infty} s_n=0\,.$ This means all elements of $S_N$ are between $\pm \varepsilon$. So $\sup S_N < \varepsilon$. Now to guarantee that all $\sup S_n$ for $n>N$ are also bounded by $\varepsilon$, I think we need the observation that $\sup S_n < \sup S_N$ for $n>N$ because $S_n\subseteq S_N$. Now the sequence $(S_n)_{n \in \mathbb{N}}$ fulfills the $\varepsilon -$criterion for convergence to zero.

I apologize, if you meant this. I think I was a bit confused. My first thought was to choose $\varepsilon = \frac{1}{N}$, but then we only have some $S_{M(N)} < \frac{1}{N}$ and it is not immediately obvious, that the convergence of this subsequence $(S_{M(N)})_N$ implies the convergence of $(S_N)_N$ as a whole. E.g. $a_n=(-1)^n$ has also convergent subsequences, but doesn't converge as a whole.

Have you tried the other direction in the meantime?

 

[fishturtle1]

I'm not sure, if I make it too complicated or not. Say $S_N := \{s_n\, : \,n>N\}$ and $\varepsilon >0$. Then there is a $N$ with $|s_n|<\varepsilon$ for all $n>N$, if we assume $\lim_{n \to \infty} s_n=0\,.$ This means all elements of $S_N$ are between $\pm \varepsilon$.

Ok i see how you got here,

So $\sup S_N < \varepsilon$. Now to guarantee that all $\sup S_n$ for $n>N$ are also bounded by $\varepsilon$, I think we need the observation that $\sup S_n < \sup S_N$ for $n>N$ because $S_n\subseteq S_N$. Now the sequence $(S_n)_{n \in \mathbb{N}}$ fulfills the $\varepsilon -$criterion for convergence to zero.

Why does $S_n\subseteq S_N$ mean that we need the observation that $\sup S_n < \sup S_N$ for $n>N$?

What would prove $\limsup s_n = 0$? I think we need to take a sequence of all $sup (s_n : n > k)$ and then see where their limit is, but how?

Have you tried the other direction in the meantime?

I have a theorem from a previous chapter:
Let $(s_n)$ be a sequence in $\mathbb{R}$.
i) If $\lim s_n$ is defined[ as a real number, $+\infty, -\infty$], then $\liminf s_n = \lim s_n = \limsup s_n$.
ii) If $\liminf s_n = \limsup s_n,$ then $\lim s_n$ is defined and $\lim s_n = \liminf s_n = \limsup s_n$.

So to show the other direction,
Suppose $\limsup \vert s_n \vert = 0$. We want to show $\liminf \vert s_n \vert = 0$. This seems reasonable because $\vert s_n \vert \ge 0$ and since the $\sup (s_n : n > k)$ keeps getting closer and closer to 0, then $\vert s_n \vert$'s should be getting closer and closer to 0, but can't get smaller than 0.

 

[fresh_42]

Why does $S_n\subseteq S_N$ mean that we need the observation that $\sup S_n < \sup S_N$ for $n>N$?

For $n > N$ we have $S_n \subseteq S_N$, because $S_N = S_n \cup \{s_{N+1},s_{N+2}, \ldots , s_{n-1},s_{n}\}$ so the supremum of $S_N$ is eventually greater than the supremum of $S_n$ in case the set $\{s_{N+1},s_{N+2}, \ldots , s_{n-1},s_{n}\}$ contains a greater element. This is the observation. Needed is it, because we have $a_N := \sup S_N < \varepsilon$, but we need $a_n < \varepsilon$ for all $n> N$. As $a_n \leq a_N < \varepsilon$ this is given.

What would prove $\lim \sup s_n = 0$?

What I wrote above: For all $\varepsilon >0$ there is an $N \in \mathbb{N}$ such that for all $n > N$ we have $|a_n| < \varepsilon$ where our sequence $(a_n)_{n \in \mathbb{N}} = (\sup S_n)_{n \in \mathbb{N}}$. You see, that the many equal letters $S_n$ for sets and sequence elements is a bit confusing. At least I read your $\lim \sup s_n = 0$ as a typo for $\lim \sup S_n = 0$. Otherwise I don't know why you want to consider $\sup s_n$. All that matters is $\lim s_n = 0$ in this part of the proof.

I think we need to take a sequence of all $sup (s_n : n > k)$ and then see where their limit is, but how?

What do the round parentheses now mean? We need $a_n = \sup S_n = \sup \,\{s_m\, : \,m>n\}$.

 

[fresh_42]

I have a theorem from a previous chapter:
Let $(s_n)$ be a sequence in $\mathbb{R}$.
i) If $\lim s_n$ is defined[ as a real number, $+\infty, -\infty$], then $\liminf s_n = \lim s_n = \lim \sup s_n$.
ii) If $\lim \inf s_n = \lim \sup s_n,$ then $\lim s_n$ is defined and $\lim s_n = \lim \inf s_n = \lim \sup s_n$.

So to show the other direction,
Suppose $\lim \sup \vert s_n \vert = 0$. We want to show $\lim \inf \vert s_n \vert = 0$. This seems reasonable because $\vert s_n \vert \ge 0$ and since the $\sup (s_n : n > k)$ keeps getting closer and closer to 0, then $\vert s_n \vert$'s should be getting closer and closer to 0, but can't get smaller than 0.

Shouldn't this be a little more formal than this? I mean, intuitively it is obvious anyway, so a couple of "closer" would not convince me.
Considering this lemma, I wonder what should be shown at all?

 

[fishturtle1]

Shouldn't this be a little more formal than this? I mean, intuitively it is obvious anyway, so a couple of "closer" would not convince me.
Considering this lemma, I wonder what should be shown at all?

Yes I meant to say that was my intuition and I couldn't think how to put this into an actual argument.

$(\rightarrow)$ Suppose $\limsup \vert s_n \vert = 0$. Let $\varepsilon > 0$. By definition of limit, there exists $N$ such that $n > N$ implies $\vert \sup \vert s_n \vert \vert < \varepsilon$. By definition of supremum and absolute value, we have $\vert s_n \vert \le \sup \vert s_n \vert \le \vert \sup \vert s_n \vert \vert < \varepsilon$. So $\vert s_n \vert < \varepsilon$. So for all $\varepsilon > 0$, there exists an N such that $n > N$ implies $\vert s_n \vert < \varepsilon$. By definition of limit, $\lim s_n = 0$.

$(\leftarrow)$ Suppose $\lim s_n = 0$. Let $\varepsilon_1 > 0$. By definition of limit, there exists $N_1$ such that $n > N_1$ implies $\vert s_n \vert < \varepsilon_1$. So $\vert \vert s_n \vert \vert = \vert s_n \vert < \varepsilon_1$. Therefore $\vert \vert s_n \vert \vert < \varepsilon_1$. Therefore, for all $\varepsilon_1 > 0$ we have $n > N_1$ implies $\vert \vert s_n \vert \vert < \varepsilon_1$. By definition of limit $\lim \vert s_n \vert = 0$. By theorem 10.7, $\lim \sup \vert s_n \vert = 0$.
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