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I aim to show that the sequence $x_n := n^2 - 10n $ diverges to $+\infty$ by using the definition of divergence (i.e. for a given $M \in \mathbb{R}$, there exists $N$ such that $n \geq N$ implies $x_n > M$).

So my strategy for proving this is that if you give me $M$, I will give you $N$ such that $n \geq N$ implies $x_n > M$.

Somehow it is getting harder than I thought it was, though. I have tried to set $N = M+10$, but I realized that that does not make sense because $M \in \mathbb{R}$ and $N \in \mathbb{N}$.

Any suggestions?

graydad
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    You could always say something like "select an integer $N \geq M + 10$". We generally take for granted that such an integer exists (the "Archimedian property" of $\Bbb R$). – Ben Grossmann Sep 15 '15 at 02:57

2 Answers2

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Starting by taking any $M > 0$, we have $n^{2} - 10n = n(n-10) > M$ if $n > M+10$, so taking $N := \lceil M+10 \rceil$ suffices. Here $\lceil a \rceil$ denotes the least integer upper bound of $a$ for all $a \in \mathbb{R}$.

Yes
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  • This has not dealt with the fact that $M \in \mathbb{R}$. We are not proving convergence so $M$ does not have to be a positive number. –  Sep 15 '15 at 03:40
  • @SeojunHong: Did you notice that $n \in \mathbb{N}$? Moreover, I think the OP's problem lies in the integer-taking function. – Yes Sep 15 '15 at 03:41
  • What does $n \in \mathbb{N}$ has to do with the fact that we have considered some $m \in \mathbb{R}$? What if $M=-2$? Then $9 > 8$, but $9 (9 -10) \ngeq -2$. –  Sep 15 '15 at 03:45
  • Well, because that means that the sequence of interest has only finitely many negative terms. @SeojunHong – Yes Sep 15 '15 at 03:46
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Let $M>0$ be given. Choose $N>1$ such that $N-10>M$. Clearly $$\frac{M}{\left|N-10\right|}<1$$ and by choice of $N$ this means $$\frac{M}{\left|N-10\right|}<N$$ It follows for all $n \geq N$ that $\frac{M}{\left|n-10\right|}<n$ and so $$\begin{align}M<n\left|n-10\right| \\ =\left|n^2-10n\right|\end{align}$$ which shows that $x_n$ diverges.

graydad
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  • This has not dealt with the fact that M∈R. We are not proving convergence so M does not have to be a positive number –  Sep 15 '15 at 03:40
  • @SeojunHong I disagree; The way the proof is written it leaves $M$ general enough to be anything. Second, $n^2-10n$ clearly tends to positive infinity. There is no need to consider any negative values of $M$. Lastly, I suggest you review the definition of convergence again. The standard definition is that $|x_n-x|<\varepsilon$ for all $\varepsilon>0$ and $n \geq N$, which means to diverge we'd need $|x_n-x| \geq M$ for all $M,x \in \mathbb{R}$ and some $n > N$. Given that the proof is done with absolute value, we necessarily require $M$ to be positive. – graydad Sep 15 '15 at 03:44
  • Well, it is the definition of divergence, so we cannot just prove one-sidedly for $M > 0$ and say there is not need to consider $M \leq 0$. I do not see that the conclusion holds clearly for $M \in \mathbb{R}$. –  Sep 15 '15 at 03:51
  • @SeojunHong Choose any $M$ you want. Seriously. It doesn't matter. Beacause I know how to find an $N$ such that $M<\left|n^2-10n\right|$ whenever $n \geq N$. This is immediately true for all $M \leq 0$ as $\left|n^2-10n\right|>0$ for all $n \in \Bbb{N}\setminus{10}$. – graydad Sep 15 '15 at 03:57
  • Yes, I see that. Thank you for your sharing. I really appreciate it. –  Sep 15 '15 at 04:14