I have this assignment:

$$\lim_{n\to \infty }\frac{(-1)^n-3}{n^2}=0$$

I have to prove that limit by definition, but I am stuck with

$$|a_n - 0|< \epsilon$$

Dont know how to deal with that absolute value.

The solution is


  • $\begingroup$ Have you tried to see what the numerator is for the first few $n$? That might give you an idea of how to deal with the absolute value. Note that the denominator is always positive. $\endgroup$ – Theoretical Economist Oct 11 '19 at 19:19

We have that

$$\frac{-2}{n^2}\le\frac{(-1)^n-3}{n^2}\le \frac{-4}{n^2} $$

and then

$$\frac{2}{n^2}\le\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} $$


$$\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} <\epsilon \implies n^2 > \frac{4}{\epsilon} \implies n>\frac{2}{\sqrt\epsilon}$$

therefore it suffices to take

$$n_0=\overbrace{\color{blue}{\left[\frac{2}{\sqrt\epsilon}\right]}}^{\color{red}{\text{integer part}}}+1$$

  • 1
    $\begingroup$ Since we need $n>\frac{2}{\sqrt\epsilon}$ we take the integer part of that and add +1. $\endgroup$ – user Oct 11 '19 at 19:28
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    $\begingroup$ @naruto25 By definition of limit we are looking for $n_0 \in \mathbb N$ but $\frac{2}{\sqrt\epsilon} \in \mathbb R$ therefore we need to extract the integer part and then add $1$. Try with a numerica example: suppose we obtain $n> 100.45$ then we take $n_0=100+1$. $\endgroup$ – user Oct 11 '19 at 19:50
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    $\begingroup$ @naruto25 We are taking an upper bound that is $4/n^2$, we don't need the exact bound but a simple upper which works. For example we also could take $6/n^2$ and the proof works at the same way. $\endgroup$ – user Oct 11 '19 at 19:52
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    $\begingroup$ @naruto25 We are thinking $\epsilon$ as a small number, for example $\epsilon=.002 \implies \sqrt \epsilon =0.04472...$ and $\frac{2}{\sqrt\epsilon}>44.72...$ then we can take $n_0=45$. $\endgroup$ – user Oct 11 '19 at 20:01
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    $\begingroup$ @naruto25 Exactly! $\endgroup$ – user Oct 11 '19 at 20:02

Given $\epsilon>0 $, we should find $N$ such that if $n\ge N$ then $$|\frac{(-1)^n-3}{n^2}|<\epsilon$$


$$|\frac{(-1)^n-3}{n^2}|\le |\frac{(-1)^n}{n^2}|+|\frac{3}{n^2}|\le \frac{4}{n^2}$$

thus it is sufficient to find one $N$ such that,

$$n\ge N \;\; \implies \;\; \frac{4}{n^2}<\epsilon$$


$$n\ge N \;\; \implies \;\; n^2 > \frac{4}{\epsilon}$$

or $$n\ge N \;\; \implies \;\; n > \frac{2}{\sqrt{\epsilon}}$$

So, each $N$ satisfying $N>\frac{2}{\sqrt{\epsilon}}$ will work.

the smallest of these $N$ is

$$\lfloor \frac{2}{\sqrt{\epsilon}} \rfloor +1$$

  • $\begingroup$ So if I had epsilon=1000000 and then it was 0. I do not have null member of that sequence and thus I have to add that 1? $\endgroup$ – naruto25 Oct 11 '19 at 19:33
  • $\begingroup$ If $\epsilon=1000000$ then $N=1$ $\endgroup$ – hamam_Abdallah Oct 11 '19 at 19:35
  • $\begingroup$ In general, $\epsilon$ Must be Small. Say if $\epsilon =0.000001$ then $N =?$ $\endgroup$ – hamam_Abdallah Oct 11 '19 at 19:37
  • $\begingroup$ I am still curious why do we have to add that one to that solution. If I had epsilon=1, then the first member of sequence(n0) from which that limit starts should be 2. And every next member should fit in distance epsilon. $\endgroup$ – naruto25 Oct 11 '19 at 19:40
  • $\begingroup$ @naruto25 If it is true for small epsilons, it will be also true for greater epsilons. $\endgroup$ – hamam_Abdallah Oct 11 '19 at 19:41

Notice, that:

$$\frac{(-1)^n-3}{n^2} \geq \frac{-4}{n^2}$$

And is always negative. So for every $\epsilon$ you can find $N$ ($=\frac{2}{\sqrt{\epsilon}})$, that for every $n>N$ the following holds: $$|\frac{(-1)^n-3}{n^2}| \leq \epsilon$$


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