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.


2 Answers 2


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 $\limsup_n (c_n+b_n) = c + \limsup_n b_n$.)

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

Here is a possible answer based on the properties

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


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

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.

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

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