# Prove that a series is divergent, with $\epsilon,N$

I have this question:

Prove that $$a_n=\frac{(-1)^nn+1}{n+2}$$ is divergent, without proving it by contradiction, or by using any theorems from the book. Prove it by using $$\epsilon$$ and $$N$$ notation.

In the book there's a sentence that $$(-1)^n$$ is divergent, and its proof, so I tried my way by manipulating it a bit.

Here is my attempt:

We need to prove that for every L exists $$\epsilon > 0$$ such for every N exists $$n>N$$ such as $$|a_n-L|\geq \epsilon$$.

Suppose there exists $$L\geq0$$, and exists $$\epsilon=1/2$$ , then: $$\frac{-n+1}{n+2} < -\frac{1}{2} \Leftrightarrow -2n+2<-n-2 \Leftrightarrow -n<-4 \Leftrightarrow n<4$$.

Now, suppose exists N, such as $$n>4\geq N$$, (No idea how to proceed..)

And now the other side, Suppose there exists $$L<0$$, and exists $$\epsilon=1/2$$ , then: $$\frac{n+1}{n+2} > \frac{1}{2} \Leftrightarrow 2n+2>n+2 \Leftrightarrow n > 0$$

I have no idea how to continue it, Thank you for your help!

Consider the subsequences $$~a_1,a_3,a_5,\cdots~$$ and $$~a_2,a_4,a_6,\cdots~$$ and (for the moment) assume that these subsequences converge to $$~-1~$$ and $$~1~$$ respectively.

The problem may then be attacked as follows:

• First prove that if the assumption is accurate, that then the original sequence $$~a_1,a_2,\cdots,~$$ can not be convergent.

• Then prove that the assumption is true.

Per the assumption, for any $$~\epsilon_1 > 0~$$ there exists a positive integer $$~N_1,~$$ such that for all $$~n \geq N_1,~$$

$$| ~a_{2n-1} - (-1) ~| < \epsilon_1. \tag1$$

Similarly, there exists a positive integer $$~N_2,~$$ such that for all $$~n \geq N_2,~$$

$$| ~a_{2n} - (1) ~| < \epsilon_1. \tag2$$

Set $$~M = \max(N_1,N_2).~$$

Then, the inequalities in (1) and (2) above must hold for any $$~n \geq M,~$$ and any $$\epsilon_1 > 0.$$ In particular, the inequalities hold for $$~\epsilon_1 = (1/10).$$

That is, when $$~\epsilon_1 = (1/10),~$$ the appropriate positive integer $$~M~$$ can be found.

This implies that for any $$n~\geq M,$$ that

• $$a_{2n-1} < (-1) + \epsilon_1 = (-0.9).$$

• $$a_{2n} > 1 - \epsilon_1 = (0.9).$$

• Therefore, $$~|a_{2n} - a_{2n-1}| > 2 - 2\epsilon_1 = 1.8.$$

Now, to prove that the (overall) sequence $$~a_1,a_2,\cdots,a_n~$$ is not convergent, it is sufficient to prove that for the specific value $$~\epsilon_2 = (1/2),~$$ it is not the case that there exists any real number $$~L~$$ such that :

there exists a positive integer $$~N~$$ such that
for any $$~n > N,~ |a_n - L| < (1/2).$$

This portion of the analysis will use the bullet pointed inequalities above to establish that it is impossible for such an $$~L~$$ to exist.

Assume that such an $$~L~$$ exists.

Then, from the bullet pointed inequalities, there must exist a positive integer $$~n~$$ such that all of the following are true:

• $$|a_{2n-1} - L| < (1/2).$$
• $$|a_{2n} - L| < (1/2).$$
• $$|a_{2n-1} - a_{2n}| > 1.8.$$

The above three bullet point inequalities can not all be satisfied. This is because the first two inequalities would imply that

$$|a_{2n-1} - a_{2n}| < |a_{2n-1} - L| + |L - a_{2n}| < (1/2) + (1/2) = 1,$$

by the triangle inequality.

Therefore, the assumption that such an $$~L~$$ exists, for the specific value $$~\epsilon_2 = (1/2)~$$ generates a contradiction. Therefore, based on the (as yet unproven assumption), the overall sequence can not be convergent.

Therefore, the entire problem is reduced to proving that the assumption is true.

Let $$~b_1,b_2,b_3,\cdots~$$ denote the subsequence $$~a_1,a_3,a_5,\cdots,~$$ and
let $$~c_1,c_2,c_3,\cdots~$$ denote the subsequence $$~a_2,a_4,a_6,\cdots.~$$

Then

$$b_n = -\dfrac{(2n-1)+1}{(2n-1)+2} = (-1) \times \left[ ~1 - \frac{1}{2n+1} ~\right].$$

I need to prove that the above sequence converges to $$~-1.$$

For any $$~\epsilon > 0,~$$ choose $$~N \in \Bbb{Z^+}~$$ such that $$~(2N+1) > \dfrac{1}{\epsilon}.~$$

This implies that for any $$~n \geq N:$$

$$(2n + 1) \geq (2N+1) > \frac{1}{\epsilon} \implies \frac{1}{2n+1} < \epsilon \\ \implies 1 > \left[1 - \frac{1}{2n+1}\right] > 1 - \epsilon \implies \\ -1 < (-1) \times \left[1 - \frac{1}{2n+1}\right] < -1 + \epsilon \implies \\ -1 < b_n < -1 + \epsilon \implies \\ |~b_n - (-1)~| < \epsilon.$$

So, the first portion of the assumption has been proven: the subsequence of odd elements does in fact converge to $$~-1.$$

$$c_n = \dfrac{2n+1}{2n+2} = \left[ ~1 - \frac{1}{2n+2} ~\right].$$

I need to prove that the above sequence converges to $$~+1.$$

For any $$~\epsilon > 0,~$$ choose $$~N \in \Bbb{Z^+}~$$ such that $$~(2N+2) > \dfrac{1}{\epsilon}.~$$

This implies that for any $$~n \geq N:$$

$$(2n+2) \geq (2N+2) > \frac{1}{\epsilon} \implies \frac{1}{2n+2} < \epsilon \\ \implies 1 > \left[1 - \frac{1}{2n+2}\right] > 1 - \epsilon \implies \\ 1 - \epsilon < c_n < 1\implies \\ |~c_n - (+1)~| < \epsilon.$$

So, the second portion of the assumption has been proven: the subsequence of even elements does in fact converge to $$~+1.$$

"Suppose there exists $$L≥0$$ and exists $$ϵ=1/2$$" Is very weird formulated. What you are want to show here is that you choose any $$L\ge 0$$, and you are trying to show that for $$\varepsilon=\frac12$$ the formula $$\forall N\exists n>N\; |a_n-L|<\varepsilon$$ holds. But for some reason you didnt write the $$L$$ (i suppose you assumed it's zero), and you consider wheter for odd $$n$$'s (because then $$a_n=\frac{-n+1}{n+2}$$), $$a_n<-\varepsilon$$. But it's not what you have to prove. Like the inequality isn't what we want. If inequality that we want to show will works for odd $$n$$'s then of course it's ok. But we want inequality $$a_n\leq L-\varepsilon$$. Of course for any bigger $$L$$ the inequality holds but you didn't mentioned that so proof isn't full.

Now, you want to take any $$L<0$$ (still, any, not that it works for some $$L$$) and you again put $$\varepsilon=\frac12$$. And in this case, well you here again assumed that $$L=0$$ and wanted to show (for some infinite amount of $$n$$'s, the even ones) $$a_n>\varepsilon$$. But you want to show that $$a_n>L+\varepsilon$$ (or $$a_n,but this case we can ommit as $$a_n$$ is positive for even's). So you have to show that $$a_n>L+\varepsilon$$. For $$L<0$$ and even $$n$$ it's true as it's equivalent to $$n>2L$$, but this inequality is always true becaues $$n>0>2L$$ for any $$n$$.

I would suggest to be very careful in what you want to prove and how, and writing clearly explaining your steps. Especially in harder things it can be very hard to understand what you were doing and how. While making prove it should be clear what the person that proves theorem is doing at the moment, and here it isn't (I'm not writing it to be mean, just to tell to pay more attention to how your proof's looks like to make better proofs in future).

So basically your proof in form as you written, but corrected to be more valid should looks like this:

You want to show something for any L, so you can consider separetely case when $$L\ge 0$$ and $$L<0$$ (as you did). Then you show for some $$\varepsilon$$ (here you choosed $$1/2$$, that is correct), and for infinietely many $$n$$'s (i write infinitely many $$n$$'s as it is what basically $$\forall N\exists n>N \;...$$ means) the inequality $$|a_n-L|\ge\varepsilon$$ holds (of course $$|a_n-L|<\varepsilon$$ is equivalent to $$a_n\le L-\varepsilon\vee a_n\ge L+\varepsilon$$, which you halfheartedly use (because the case of forgotting the $$L$$) ). Showing that for all odd (or even in second case or any other infinite subset of natural numbers) numbers our inequality works is telling that there are infinitely many n's such that this thing works, so after that is just showing the inequality. Here we can do this simmilary to what you did but put the L next time and at least describe a little what you're doing there not just write some arbitrary inequality with some arbitrary expressions with just showing that this arbitrary stuff has some arbitrary solution that you not use in describing what does it mean for proof.