Lim sup(an+bn) less than or equal to

[spenghali]

Homework Statement

If limsup(an) and limsup(bn) are finite, prove that limsup(an+bn) \leq limsup(an) + limsup(bn).

Homework Equations

The Attempt at a Solution

My proof seems a bit short, so if someone could please reassure me this is a valid proof, thanks in advance.

Proof: Assuming an and bn are bounded sequence. Let a > limsup(an) and b > limsup(bn). Then a+b > an+bn for all but finitely many n's. This implies that a+b \geq limsup(an+bn). Since this hold for any a \geq limsup(an) and any b > limsup(bn), this implies limsup(an+bn) \leq limsup(an) + limsup(bn). QED

 

[lanedance]

looks reasonable to me (though i am rusty at this)

so just to sum up

an > limsup(an), bn > limsup(bn) for only finitley many n

an + bn > limsup(an + bn) for only finitley many n

which implies
limsup(an + bn) <= limsup(an) + limsup(bn)

 

[spenghali]

thanks for the input, unfortunately my professor refused to tell me if i was right.

 

[some_dude]

If limsup(an) and limsup(bn) are finite, prove that limsup(an+bn) \leq limsup(an) + limsup(bn).

...

Proof: Assuming an and bn are bounded sequence. Let a > limsup(an) and b > limsup(bn). Then a+b > an+bn for all but finitely many n's. This implies that a+b \geq limsup(an+bn). Since this hold for any a \geq limsup(an) and any b > limsup(bn), this implies limsup(an+bn) \leq limsup(an) + limsup(bn). QED

Your first sentence is too strong an assumption. E.g., let x_n = -n, if n even, 1 otherwise. Then lim sup x_n = 1, but the sequence {x_n} is not bounded.

Edit: oops, I just looked at your proof, you didn't even use the boundedness...