Show that certain lim sup additivity implies convergence. Code Answer

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I’m trying to prove the following:

Let ${a_n}$ be a bounded sequence. If for every bounded sequence ${b_n}$ the following holds:

$$1) limsup_{n to infty} (a_n + b_n) = limsup_{n to infty} a_n + limsup_{n to infty} b_n$$

Show that ${a_n}$ is convergent.

Outline of Proof:

Here’s the outline of my proof. But there’s a point I get stuck below. I will be using the following theorem:

2) Let ${a_n}$ be a bounded sequence, let $L = limsup_{n to infty} a_n$. If $epsilon > 0$ there exists infinitely many $n$ such that $a_n > L – epsilon$, and there exists $N_1 in mathbb{N}$ such that if $n ge N_1$ then $a_n < L + epsilon$

Since condition $1)$ holds for every bounded sequence ${b_n}$ we can assume it holds for a ${b_n}$ that’s convergent. That means $forall epsilon > 0$ $exists N_2 in mathbb{N}$ $ni$ if $n ge N_2$ then $L_b – epsilon < b_n < L_b + epsilon$ where $L_b = lim_{n to infty} b_n$.

Let $epsilon > 0$, $L_a = limsup_{n to infty} a_n$, $L_b = limsup_{n to infty} b_n$. Let N = $max{N_1, N_2}$.

If $n ge N$ we can easily show $a_n < L_a + epsilon$.

This is the point where I’m stuck. For infinitely many $n$:

$$a_n + b_n > L_a + L_b – epsilon/2$$
$$a_n > L_a + L_b – b_n – epsilon/2$$

$forall n ge N$: $L_b – b_n > -epsilon/2$.

But the only conclusion I can draw from the above is that for infinitely many $n$ (and not $forall n ge N$):

$$a_n > L_a – epsilon$$


Prove that
limsup(-a_n) = -liminf(a_n).
Then with setting $b_n := -a_n$ you are done.

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