Prove that the sequence is divergent using $\varepsilon $ , $N$ definition. $a_n=\frac{(-1)^{n}n+1}{n+2}$

I am still new to this topic but what I tried :

we need to prove that for every $L$$\in R$ exists $\varepsilon \gt0$ such that for every $N \in \Bbb N$ there exists $n \gt N$ that satisfies $|a_n - L| \geq \varepsilon $.

since we have $(-1)^n$ then we need to seperate to two cases , one case where $L \lt 0$ and one where $L \geq 0$

we also need to choose a $\varepsilon$ that fulfills that so since the sequence behaves according to $\frac{(-1)^{n}n}{n}$ our radius is $-1,1$ since the sequence elements are next to $-1$ and $1$ then I choose $\varepsilon=\frac{1}{2}$ now I cannot figure out how to continue and how to prove $|a_n - L| \geq \varepsilon $ , I dont have any $L$ I only know that it can be positive or negative and I know that I have to choose my $n$ accordingly meaning $n=2N$ or $n=2N+1$

I feel lost in this topic I am trying to practice according to the rules but I feel like I am lacking understanding in it. and how should I continue from here?

Thanks for any help and tips!


1 Answer 1


Addendum just added to respond to the comment/questions of Adamrk.


$$a_n = (-1)^n \left[1 - \frac{1}{n+2}\right].\tag1$$

At this point, the rewrite of $a_n$ informally suggests that the odd elements in the sequence go towards $(-1)$ while the even elements in the sequence go towards $(+1)$.

So, this is the point where you have to use this insight to plan your attack. Then, you use formal steps to implement the attack.

One major problem, in demonstrating that the sequence is divergent is that the supposed limit $L$ is unspecified. This means that you have to demonstrate that regardless of what value of $L$ is chosen, it is impossible for the sequence to converge to that value $L$.

This problem will be handled as follows:

Set $\epsilon = (1/2)$.


Suppose that it can be shown that regardless of how large $N \in \Bbb{Z^+}$ is taken, there will exist distinct values $n_1$ and $n_2$, each $> N$, such that $|a_{n_1} - a_{n_2}| > 2\epsilon = 1.$

Then, by the triangle inequality, regardless of what fixed value $L$ is chosen, you have that

$|L - a_{n_1}| + |L - a_{n_2}| \geq |a_{n_1} - a_{n_2}| > 2\epsilon.$

This means that regardless of what value of $L$ is chosen, at least one of the two terms, $|L - a_{n_1}|$ or $|L - a_{n_2}|$ must be $> \epsilon$.

This will imply that the sequence does not converge to $L$. Further this analysis is independent of what value of $L$ is chosen, which implies that the sequence is divergent.

Therefore, the problem has been reduced to formally proving that Supposition-1 is accurate.

Given the rewrite in (1) above, let $N$ be any positive integer $> 10$.
I then have to show that regardless of how large $N$ is,
there exists distinct $n_1, n_2 > N$ such that Supposition-1 is satisfied.

From (1) above, it is immediate that the sequence of absolute values,
$|a_1|, |a_2|, |a_3|, \cdots$,
is strictly increasing,
since the fraction $\displaystyle \frac{1}{n+2}$ is strictly decreasing.

Given $N$, choose $n_1$ to be any odd integer $> N$.
Then, you have that

  • $\displaystyle 1 > |a_{n_1}| > |a_N| \geq |a_{10}| = 1 - \frac{1}{12}$.
  • $a_{n_1}$ is negative, because of the $(-1)^n$ factor.

Therefore, you have that $\displaystyle -1 < a_{n_1} < \frac{-11}{12}$.

Then, choose $n_2$ to be any even integer $> N$.
Then, you have that

  • $\displaystyle 1 > |a_{n_2}| > |a_N| \geq |a_{10}| = 1 - \frac{1}{12}$.
  • $a_{n_2}$ is positive, because of the $(-1)^n$ factor.

Therefore, you have that $\displaystyle 1 > a_{n_2} > \frac{11}{12}$.

with $~\displaystyle a_{n_1} < \frac{-11}{12}$
and $~\displaystyle a_{n_2} > \frac{11}{12}$

you must have that $|a_{n-2} - a_{n_1}| > 1 = 2\epsilon.$

Thus, Supposition-1 is established, so the sequence is proven to be divergent.

Responding to the comment questions of Adamrk.

In order to understand my motive for focusing on Supposition-1, re $|a_{n_1} - a_{n_2}| > 1$, you first have to understand :

  • What the $\epsilon$ definition is of a sequence being convergent.

  • What is entailed in negating that $\epsilon$ definition in order to show that the sequence is not convergent (AKA the sequence is divergent).

The $\epsilon$ definition of a convergent sequence is:

  • There exists a (finite) limit $L$

  • Such that for all $\epsilon > 0$

  • There exists an $N_\epsilon \in \Bbb{Z^+}$
    where the notation $N_\epsilon$ emphasizes that $N$ depends on $\epsilon$

  • Such that for all $n\in \Bbb{Z^+}$ such that $n \geq N_\epsilon$

  • $|a_n - L| < \epsilon.$

So, the above $\epsilon$ definition of convergence of a sequence is $5$ layers deep.

In order to show that a sequence is divergent, rather than convergent, you have to show that the negation of the above $\epsilon$ definition is true. This means that you have to show that:

  • For any (proposed) finite limit $L$

  • There exists at least one value for $\epsilon > 0$

  • Such that no matter how large you take $N \in \Bbb{Z^+}$

  • There will exist at least one $n \in \Bbb{Z^+}$ such that $n \geq N$

  • And such that $|a_n - L| \geq \epsilon.$

So, if you examine the formal $\epsilon$ negation of convergence, you see that given any proposed limit $L$, one is permitted to choose an $\epsilon$, based on this value of $L$ such that the above negation holds.

Instead of having $\epsilon$ depend on the value of the proposed limit $L$, I took the (valid) shortcut of arbitrarily setting

$\displaystyle \epsilon = \frac{1}{2}.$

The idea is that this (fixed) value of $\epsilon$ will be used in the negation, regardless of what proposed value of $L$ is chosen. This idea simplifies proving the negation.

Then, I had to deal with the problem that I need to show that no matter what value of $L$ is (arbitrarily) considered, the sequence does not converge to that value of $L$.

Now, go back to the start of my answer, at the section that begins:


Start re-reading from that point, down to the point that ends:

Therefore, the problem has been reduced to formally proving that Supposition-1 is accurate.

This portion of my original answer, which you have just re-read, should explain the relevance of my Supposition-1. I certainly agree that the definition of convergence is convoluted, the negation of convergence is convoluted, and the relevance that I attached to Supposition-1 is convoluted.

So, at this point, the question is:

Are you clear on the relevance that I attached to Supposition-1?

If not, leave a further comment, and I will respond with an Addendum-2.

There is also the auxiliary issue that you might also with to question:

Are you clear on the validity of my proof of Supposition-1?

  • $\begingroup$ Thank you for the clear and detailed explanation , just one question about the supposition , why did you do $|a_{n_2} - a_{n_1}| \gt 1$ I know it is accordingly to the definition $|an-L| \geq \varepsilon$ but I mean the idea of $|a_{n_2} - a_{n_1}| \gt 1$ where did it come from? $\endgroup$
    – Adamrk
    Nov 7, 2021 at 9:46
  • $\begingroup$ $|a_{n_2} - a_{n_1}| $resembles the distance of the values of the sequence if I understood right but since it goes from $-1$ to $1$ isn't $d(1,-1)=2$? my question is what does it give us that it is greater than 1? thank you , and sorry for the questions as I mentioned I am having a hard time with real analysis and struggling to understand $\endgroup$
    – Adamrk
    Nov 7, 2021 at 11:05
  • 1
    $\begingroup$ @Adamrk See the Addendum that I just added to my answer. $\endgroup$ Nov 7, 2021 at 12:27
  • $\begingroup$ Wow thank you so much! for the explanation , you proved it in a way where did not really need $L$ because it is true for all its values you just proved the inequality $|a_{n_2}-a_{n_1}| \gt 1$ In the book they solved it differently they separated into 2 cases one case where $L \lt 0$ and the other $L \geq 0$ .. for $L \lt 0$ they checked when $\frac {-n+1}{n+2} \lt -0.5$ after solving they got $n \gt 4$ then they solved $|a_n - L| =L -a_n \geq -a_N \geq -0.5$ for even $n$ and check the case $L \geq 0$ $\frac {n+1}{n+2} \gt 0.5$ and did the same thing so I got confused $\endgroup$
    – Adamrk
    Nov 7, 2021 at 13:12
  • 1
    $\begingroup$ @Adamrk The triangle inequality is always valid, so the general approach of finding $a_{n_1}, a_{n_2}$ such that $|a_{n_1} - a_{n_2}| > 2\epsilon$ should always be valid. $\endgroup$ Nov 7, 2021 at 13:26

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