showing the sum of convergent and divergent sequence is divergent

k3k3

1. The problem statement, all variables and given/known data

Show that the sum of a convrgent sequence and a divergent sequence must be a divergent sequence. What can you say about the sum of two divergent sequences?

2. Relevant equations
A theorem in the book states:
Let {a_n} converge to a and {b_n} converge to b, then the sequence {a_n+b_n} converges to a+b

Definition of convergent:
A sequence {a_n} converges to a number L if for each epsilon > 0 there exists a positive integer N such that |a_n-L| < epsilon for all n ≥ N. The number L is called the limit of the sequence. The sequence {a_n} converges iuf there exists a number L such that {a_n} converges to L. The sequence {x_n} diverges if it does not converge.

3. The attempt at a solution
I think I made this too simple and overlooked something... Here is my attempt.

Let {x_n} be a sequence that converges to x.
Let {y_n} be a divergent sequence.

Suppose {x_n+y_n} is a convergent sequence.
By the theorem in the book, this implies that {y_n} converges, but {y_n} diverges.
Hence, the sum must be divergent.

What can you say about the sum of two divergent sequences? Their sum is divergent.

micromass

How does that follow from the theorem in the book?? What do you take as a_n and b_n??

Not necessarily.

k3k3

If we assume that the sum of the convergent sequence and divergent sequence is convergent, and use that the theorem the book states, both sequences must be convergent. There should be some number that {y_n} converges to but there isn't, so it can't be.

micromass

Yes, but what do you take as a_n and b_n??

k3k3

{x_n} is one of them and I was wanting to show that {y_n} is neither.

DrewD

This is a one way implication. It does not imply that if the sum is convergent then the two initial series are convergent. Note though, that the difference between two convergent series is convergent.

k3k3

Would it be safe to put both sequences on top of each other? Since they are sequences, they are in 1-1 correspondence with the natural numbers. So {x_n+y_n}=x_1,y_1,...,x_n,y_n,...

If {x_n} converges to x, but {y_n} converges to nothing, the {y_n} terms are what break the sequence from converging.