# Proving a sequence diverges directly

1. May 11, 2013

### shoeburg

I want to prove the sequence a(n) = n diverges, directly, without the aid of any theorems.
Naturally, I try to prove this by contradiction. Here's my attempt:

Let L be a real number such that a(n) converges to L. Then for all e>0, there exists a natural number N s.t. any n>N implies d(a(n) - L) < e. So I say fine, fix that N and pick e=(1/2). Pick n such that n is the next natural number that is greater than N+L. Then d(a(n) - L) = d(n - L) > d(N+L-L) = N > (1/2).

My question is did I do anything illogical by the way I picked my e and fixed the N? Also, since L is fixed, are my inequalities justified? I'm still getting used to the importance of the order of quantifiers and such.

2. May 12, 2013

### willem2

you should pick e = (1/2) first, and then conclude that there exists a number N s.t for any n>N implies d(a(n) - L) < e

If you choose a random N, it's perfectly possible that d (a(n) - L) > (1/2) for some n>N and that the function still converges. example a(n) = 10/n, N = 3

3. May 12, 2013

### shoeburg

I see your point. What if say I pick e=(1/2), then say for ANY natural number N, pick n to be the next greatest natural number after N+L. Then follow through with the same inequalities I have above. Because since then this would work for all N, there would not exist an N to counterexample my argument as you have done. Does this resolve the issue? I appreciate your helping me.