Proving the divergence of a sequence

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morson

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Given a sequence u(n) = (fn + g)/(hn + i) where f, g, h, and i are all real numbers and n belongs to the set of all natural numbers.

How do we prove that, if h = 0, that the sequence diverges?

My attempt: Give a proof by contradiction. If we assume the sequence converges to L if h = 0, then for any e > 0, there exists an N such that n > N, for which:

| (fn + g)/i - L | < e

(fn + g)/i - L < e, thus (fn + g)/i < e + L

This is where I get stuck. I think it's a matter of choosing the appropriate e to bring about a contradiction?
 
morson said:
Given a sequence u(n) = (fn + g)/(hn + i) where f, g, h, and i are all real numbers and n belongs to the set of all natural numbers.

How do we prove that, if h = 0, that the sequence diverges?

My attempt: Give a proof by contradiction. If we assume the sequence converges to L if h = 0, then for any e > 0, there exists an N such that n > N, for which:

| (fn + g)/i - L | < e

(fn + g)/i - L < e, thus (fn + g)/i < e + L

This is where I get stuck. I think it's a matter of choosing the appropriate e to bring about a contradiction?
Click to expand...

Is there any restriction on the value of f, g & i.

More specifically, if f =0 (and h = 0), then the sequence is a constant, hence convergent.
 
Yeah. Worked it out myself.

Good catch on the f = 0 and h = 0 scenario, by the way.

Thanks for the help.
 
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