Could you help me with lim inf of sequences?

Question

Can you help me prove this?

If a_n, b_n are real sequences, a, b are in R, a_n + b_n -> a +
> b, and a <= lim inf a_n, b <= lim inf b_n, then a_n -> a and b_n -> b.
.

thank you

Answer 1

I think there's a mistake in Sean's proof. His first 2 inequalities seem to be mistaken.

A possible proof is

According to the properties of sequences of real numbers, we have

lim inf(a_n + b_n) = lim (a_n + b_n) = a + b >= lim inf (a_n) + lim inf(b_n) (1).

Since a <= lim inf a_n, b (2) and b <= lim inf b_n (3), it follows (2) and (3) must be equalities, so that

a = lim inf a_n and b = lim inf b_n (3)

a_n contains a subsequence a_n_k such that lim a_n_k = lim sup a_n. So, lim b_n_k = a + b - lim sup a_n >= lim if b_n = b, so that lim sup a_n <= a. Hence, lim inf a_n = lim sup a_n = lim a_n = a, which automatically implies that lim b_n = b.

Answer 2

Consider a_n + b_n - b_n = a_n.

So

lim inf(a_n + b_n) - lim sup b_n = lim inf a_n

So

a+b - lim sup b_n >= a

implies lim sup b_n <= b

Thus b_n -> b. Same sort of thing to show a_n -> a.