# Proof verification for proving the inequality $\limsup(a_n+b_n)\ge \liminf a_n+\limsup b_n$

I want to prove that for any sequences $$(a_n)$$ and $$(b_n)$$, the following holds $$\limsup(a_n+b_n)\ge \liminf a_n+\limsup b_n \tag 1$$

I want to prove the above inequality only for the case when quantities on RHS are finite.

I'm using the following definition.

Definition: $$\liminf_{n\to \infty} x_n= \lim_{n\to \infty} (\inf_{m\ge n}x_m)$$, where quantity in parentheses on RHS is $$\inf\{x_m: m\ge n\}$$.

Now, RHS can be simplified using the definition as follows:

$$\liminf a_n+\limsup b_n=\lim (\inf_{m\ge n} a_m+\sup_{m\ge n} b_m)$$. Let $$u_n:=\sup_{m\ge n} b_m$$

For any $$n$$, there exists some $$m'\ge n$$ such that $$\inf_{m\ge n} a_m+(u_n-\frac 1n)\le a_{m'}+b_{m'}\le\sup_{m\ge n}(a_m+b_m)\tag 2$$ It follows by $$(2)$$ that $$\inf_{m\ge n} a_m+u_n-\frac 1n\le \sup_{m\ge n}(a_m+b_m)$$, which gives $$\lim_{n\to \infty}(\inf_{m\ge n} a_m+u_n-\frac 1n)\le\lim_{n\to \infty}\sup_{m\ge n}(a_m+b_m)\tag 3$$ and LHS of $$(3)$$ (by limit rules and noting that $$\frac 1n\to 0$$) simplifies to $$\liminf a_n+\lim u_n\le\limsup (a_n+b_n)$$. This proves $$(1)$$.

Is my proof correct? Thanks.

You are making it too complicated.

$$(a_n+b_n) \ge \inf_{m \ge n}a_m+b_n$$ for all $$n$$.

Hence $$\sup_{k \ge n} (a_k+b_k) \ge \sup_{k \ge n} (\inf_{m \ge k}a_m+b_k)$$.

Note that $$\inf_{m \ge n}a_m \to \liminf_n a_n = \sup_n \inf_{m \ge n}a_m$$, hence taking limits we get the desired result.

(Note that if $$c_n \to c$$, then $$\limsup_n (c_n+b_n) = c + \limsup_n b_n$$.)

• Take $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$. For $n$ even your first display does not hold. Oct 27, 2021 at 5:54
• @VáclavMordvinov I need more alcohol. Thanks for catching that. Oct 27, 2021 at 5:55

Here is a possible answer based on the properties

$$\limsup_{n \to \infty}(a_n + b_n) \leq \limsup_{n \to \infty}(a_n) + \limsup_{n \to \infty}(b_n)$$

and

$$- \limsup_{n \to \infty} (-a_n) = \liminf_{n \to \infty} (a_n).$$

Both of these follow from the definitions. Well,

$$b_n = (b_n + a_n) - a_n$$

so that the first property implies

$$\limsup_{n \to \infty}(b_n) \leq \limsup_{n \to \infty}(b_n + a_n) + \limsup_{n \to \infty}(-a_n)$$.

Now move $$\limsup_{n \to \infty}(-a_n)$$ to the LHS and use the second property.

• Is the step just before $(3)$ in my post correct or is my overall proof correct? Thanks.
– Koro
Oct 27, 2021 at 7:06