Question: $ \left(\frac{1}{n}-n\right)_{n}$ tends to $-\infty \text { as } n \rightarrow \infty$

I found a proof that I don't understand. Are there any steps missing or is it incorrect...?

Proof Given:-

Let $K \in \mathbb{R}$. Then set $N \in \mathbb{N}$ so that $N>1-K$. So $n \geqslant N$ implies $$ x_{n}=\frac{1}{n}-n \leqslant 1-N<K [QED] $$ I'm not able to understand how $N>1-K$

While attempting, I instead got the condition $\left(\sqrt{\frac{k^{2}}{4}+1}-\frac{k}{2}\right) < N$ (I can prove it if requested/needed)

It could be that my condition is the 'optimal' one and that the other is an approximation (which would make both valid as an inequality) but I'm failing to correlate the two or where to make the approximation.

  • 3
    $\begingroup$ Could you clarify what the subscript $n$ on $\left(\frac{1}{n}-n\right)$ should be taken to mean? $\endgroup$
    – Tamay
    Sep 26 '20 at 14:39
  • 1
    $\begingroup$ @Tamay The question is an example in my Prof.'s notes. He's explained it to mean "a standard sequence notation" afaik. $\endgroup$
    – user220704
    Sep 26 '20 at 14:57
  • $\begingroup$ If I understand correctly, you're able to see why such an $N$ exists, but not why we chose it? $\endgroup$ Sep 26 '20 at 14:58
  • $\begingroup$ so long as $1- K \ge \sqrt{\frac {k^2}4 + 1}-\frac k2$ you don't have any problem. $\endgroup$
    – fleablood
    Sep 26 '20 at 15:32
  • $\begingroup$ The $N$ is a choice which depends of course on $K$. For instance, you may take $N$ as the integer part of $K$ minus 1 in the proof you quoted in the OP. The $N$ need not be optimal. $\endgroup$ Sep 26 '20 at 15:34

For any $k$ let $W_k = \{N| n>N \implies \frac 1n-n < k\}$. To show that $\frac 1n -n \to -\infty$ we must show that $W_k$ is not empty. We do not have to calculate what $\min W_k$ is. What you did is you calculated what $\min W_k$ was. That took some hard calculations but it showed that $W_k$ was not empty. What your professor did was must show $1-k \in W_k$ so $W_k$ is not empty. Showing $1-k \in W_k$ was incredibly easy.

Both show that $\frac 1n-n \to -\infty$.

Note $\sqrt{\frac {k^2}4 + 1} -\frac k2 < 1-k$. So if $n > 1-k > \sqrt{\frac {k^2}4 + 1} -\frac k2$ they both imply $\frac 1n - n < K$.


No one cares about the optimal value. For every $K$ there an $N$ so that $n> N$ will imply $n-\frac 1n < K$. Which means that for any $\mathscr N >N$ we will also have $n > \mathscr N$ then $n > N$ so $n-\frac 1n < K$.

Now NOBODY cares about finding the smallest possible such $N$, because finding ANY such $\mathscr N$ will do.

Finding the smallest possible such $N$ is hard:

($\frac 1n - n) \le K$ so $n^2 +Kn -1\le 0$ so $n^2 + Kn +(\frac {K^2}4) \le 1+ (\frac {K^2}4)$ etc....$n < \sqrt{\frac {k^2}4 + 1} -\frac k2$ so if $N = \sqrt{\frac {k^2}4 + 1} -\frac k2$ we ar good.

But finding any is easy:

If $n >1$ then $\frac 1n < 1$ so $\frac 1n - n < 1 - n$ so if $1-n < K$ then $\frac 1n - n < 1-n < K$ and so if we let $\mathscr N = 1-K$ (assuming $K< 0$) then if $n > \mathscr N \implies n > 1-K \implies -n < K-1 \implies \frac 1n -n < 1 - n < K$.

Now admitted $\mathscr N$ might not be the smallest such number with the the property that $n > \mathscr N\implies \frac 1n - n < K$, and indeed $\mathscr N > N$ and $N$ also has the property, but.... who cares? No one ASKED for the smallest such number. And we don't care when $\frac 1n -n$ passes "the point of no return".

We just want to show $\frac 1n - n \to -\infty$. And we can show that with any $\mathscr N \ge N$.

  • $\begingroup$ Could you elaborate on why $$n>1$$? Also, if this is feasible, what are the implications on the proof itself? $\endgroup$
    – user220704
    Sep 26 '20 at 16:16
  • $\begingroup$ $n\to \infty$. We care what $n$ does as it gets "big". So we don't care what it does when $n < 1$. Also $n \in \mathbb N$ so $n \ge 1$. All right. Maybe I should have use $\ge$ and avoid the whole issue. But the issue doesn't matter. Had it beent $\frac 1{n-7.3} - n$ then principal is the same. We need an $N$ so that $n\ge N$ gives our result. If we have $N \ge 8$ we can say that if $n \ge N \ge 8$ then we can assume $n\ge 8$. We don't care about $n < 8$. $\endgroup$
    – fleablood
    Sep 26 '20 at 16:24

Let $x_n = \frac{1}{n}-n$ for all $ n \geqslant 1$. The objective then is to show that for any negative $K$, no matter how large, all $x_n$ after some point will be less than $K$. We use negative $K$ since we want to show $x_n \to -\infty$. That is what the statement $x_n \to -\infty$ as $n \to \infty$ means.

It amounts to saying for every $K < 0$ we can find $N$, dependent on $K$, such that, $$ x_n < K \quad\text{for all }n \geqslant N. $$ But for given $K$ this is obtained by choosing a value for $N$ greater than the positive number $1-K$. Then if $n \geqslant N$, by definition we have $n > 1 - K$ which means $-n \leqslant K-1$. At the same time, $n$ is always greater than or equal to one, so $1/n$ is less than or equal to one. We can add the two inequalities, $$ 1/n \leqslant 1, \quad -n \leqslant K-1 $$ to get $$ \frac{1}{n}-n \leq K. $$

I hope that is what you are looking for.


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