Proving fractional inequality

I have a question on series, and in the end it boils down to showing that:

[LATEX]4\leqslant \frac{7n+1}{n+1} < 7,\; n\in \mathbb{Z}^{+}[/LATEX]

Is there a simple way I can "show that"? It is easy to see by subbing numbers in. NB: It doesn't say "prove that", however I feel subbing in 1 and 1E100 isn't enough.

Hi,

If you divide the numerator by the denominator you get a simpler equation to evaluate which makes it more apparent.

Ah, thank you MrAl. That gives:

[LATEX]7-\dfrac{6}{n+1}[/LATEX]

Correct?

So

[LATEX]7-\dfrac{6}{n+1} < 7[/LATEX]

because as n -> ∞, 6/(n+1) -> 0, giving 7 - δ < 7

and

[LATEX]7-\dfrac{6}{n+1} \geqslant 4[/LATEX]

because the smallest positive integer is 1 and 7-(6/2)=4

ARandomOWl,

Multiply by n+1 to get: 4n+4 ≤ 7n+1 < 7n+7 where n+1 > 0

Subtract 1 to get: 4n+3 ≤ 7n < 7n+6

Subtract 4n to get: 3 ≤ 3n < 3n+6

Divide by 3 to get: 1 ≤ n < n+2

So any n greater than 1 will satisfy the relationship.

Ratch