# Convergence of a sequence

1. Dec 31, 2005

### Castilla

Facts:
1. We have the decrecient sequence {a_n} which converges to "a".
2. Let "K" be a constant.
3. We have the decrecient sequence {b_n} where b_n = max { a_n, K }.

I read in Apostol that {b_n} converges to max {a, K} but I can't figure how. Can you help me? Thanks.

2. Dec 31, 2005

### Muzza

Decreasing.

The most straight-forward way is probably to separately consider the cases K >= a and K < a.

3. Dec 31, 2005

### Castilla

Sorry that I can't use latex, there is some problem in the PC.

1) Let us suppose that K >= a. Then for all epsilon > 0 we can find N(e) / if n > N then l b_n - K l = l max (a_n, K) - K l < epsilon.

2) Now let us suppose that a > K. Then for all epsilon > 0 we can find N(e) / if n > N then l b_n - a l = l max (a_n, K) - a l < epsilon.

Hm, excuse me but I cant see how these statements help to prove that max (a_n, K) tends to max (a, K).

4. Dec 31, 2005

### Muzza

Now that I think about it, it's probably slightly easier to consider a >= K and a < K (not that it makes much of a difference really).

There is no need to revert to the definition of convergence. (Or at least the epsilon stuff will be hidden in well-known theorems we can use).

Suppose a < K. Eventually a_n < K. So, eventually b_n = what?

5. Dec 31, 2005

### Castilla

Two options: a >= K, K > a.

We try first a >= K. Then there is an N such that for any n > N we got a_n >= K. Then for any n > N we got b_n (equal, by definition, to max {a_n, K} ) = a_n. So b_n converges to "a", which is max {a, K}.

Now we try a < K. Then there is an N such that for any n > N we got a_n < K. Then for any n > N we got b_n (equal, by definition, to max {a_n, K} ) = K. So b_n converges to "K", which is max {a, K}.

Thank you, Muzza, happy new year for you and your beloved ones!!!