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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.