ARandomOWl New Member Dec 11, 2011 #1 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.
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.
M MrAl Well-Known Member Most Helpful Member Dec 11, 2011 #2 Hi, If you divide the numerator by the denominator you get a simpler equation to evaluate which makes it more apparent.
Hi, If you divide the numerator by the denominator you get a simpler equation to evaluate which makes it more apparent.
ARandomOWl New Member Dec 11, 2011 #3 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
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
R Ratchit Well-Known Member Dec 14, 2011 #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 Last edited: Dec 14, 2011
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