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In order to prove, using the [itex]\epsilon-N[/itex] definition, that the sequence
[tex]\left\{\frac{n^2}{n^2+4}\right\} = \left\{\frac{1}{1+\frac{4}{n^2}}\right\}[/tex]
converges towards 1, I did the following: The sequence converges towards 1 if
[tex]\forall \epsilon>0, \exists N\in\mathbb{R} \ \mbox{such that} \ \forall n\in\mathbb{N}, n>N \Longrightarrow \left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\epsilon[/tex]
We have that
[tex]\left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\left|\frac{1}{1+\frac{4}{n^2}}\right|[/tex]
so if we can find the N for the creature on the right side of the inequality, it will also be true for the one on the left side. And then we solve and I spare you the following steps because my question is that if my line of reasoning is correct so far, then proving that there exist an N for the right member proves that there exists an N for
[tex]\left|\frac{1}{1+\frac{4}{n^2}}-a\right|[/tex]
where a in any positive real. So it doesn't REALLY show that the limit is 1 because should I try to prove that the limit is "a, any positive real", I would arrive to the same answer! All it shows is that the sequence converges. Right?
[tex]\left\{\frac{n^2}{n^2+4}\right\} = \left\{\frac{1}{1+\frac{4}{n^2}}\right\}[/tex]
converges towards 1, I did the following: The sequence converges towards 1 if
[tex]\forall \epsilon>0, \exists N\in\mathbb{R} \ \mbox{such that} \ \forall n\in\mathbb{N}, n>N \Longrightarrow \left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\epsilon[/tex]
We have that
[tex]\left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\left|\frac{1}{1+\frac{4}{n^2}}\right|[/tex]
so if we can find the N for the creature on the right side of the inequality, it will also be true for the one on the left side. And then we solve and I spare you the following steps because my question is that if my line of reasoning is correct so far, then proving that there exist an N for the right member proves that there exists an N for
[tex]\left|\frac{1}{1+\frac{4}{n^2}}-a\right|[/tex]
where a in any positive real. So it doesn't REALLY show that the limit is 1 because should I try to prove that the limit is "a, any positive real", I would arrive to the same answer! All it shows is that the sequence converges. Right?
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