Post History
#4: Post edited
by
Snoopy
·
2024-07-17T13:26:13Z (5 months ago)
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- - (2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
- - (3) One obvious direction of the inequality is proved by definition: $$
- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
&=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\&=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- - (2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
- - (3) One obvious direction of the inequality is proved by definition: $$
- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)&\text{linearity}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)&\text{by fact (2)}\\
- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
#3: Post edited
by
Snoopy
·
2024-01-09T01:37:45Z (12 months ago)
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- - (2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
- (3) $$- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\
- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- - (2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
- - (3) One obvious direction of the inequality is proved by definition: $$
- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\
- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
#2: Post edited
by
Snoopy
·
2024-01-09T01:35:16Z (12 months ago)
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- (2) If the limit of $x_n$ as $n\to\infty$ exists, then $\liminf_nx_n=\limsup_nx_n=\lim_nx_n$.- - (3) $$
- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\
- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- We will use the following facts in the proof:
- - (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- - (2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
- - (3) $$
- \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
- **Proof.**
- $$
- \begin{align}
- \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\
- &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\
- \end{align}
- $$
- which implies that
- $$
- \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
- $$
#1: Initial revision
by
Snoopy
·
2024-01-09T01:34:45Z (12 months ago)
We will use the following facts in the proof:
- (1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
- (2) If the limit of $x_n$ as $n\to\infty$ exists, then $\liminf_nx_n=\limsup_nx_n=\lim_nx_n$.
- (3) $$
\liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
$$
**Proof.**
$$
\begin{align}
\liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\
&=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\
&=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\
&=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\
\end{align}
$$
which implies that
$$
\liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)
$$