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Prove whether the following sequence converges or diverges: $a_n$=$\frac{(-1)^{n}n}{n+1}$

Claim: the sequence diverges

Proof:

Consider the subsequence $\frac{(-1)^{2n}2n}{2n+1}$ of $(a_n)$

Then, $\lim\frac{(-1)^{2n}2n}{2n+1}=1$ as $n \to \infty$

Let $\epsilon>0$ be given. Choose $N\geq \frac{1}{\epsilon}$. Then, for all $n\geq N$, $|\frac{(-1)^{2n}2n}{2n+1}-1|<\epsilon$

Consider another subsequence $\frac{(-1)^{2n-1}(2n-1)}{2n}$ of $(a_n)$

Then $\lim\frac{(-1)^{2n-1}(2n-1)}{2n}=-1$ as $n \to \infty$

Let $\epsilon>0$ be given. Choose $N\geq \frac{1}{\epsilon}$. Then, for all $n\geq N$, $|\frac{(-1)^{2n-1}(2n-1)}{2n}-(-1)|<\epsilon$

Since there are two subsequences of $a_n$ converging to two different limits, then, this implies that the sequence $(a_n)$ diverges.

Can anyone please verify if the following proof is correct? Also, are there any other ways to prove divergence?

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  • $\begingroup$ $\lim a_n$ doesn't exist hence the sequence diverges. $\endgroup$ – kingW3 Mar 22 '18 at 19:24
  • $\begingroup$ @A.Asad Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 24 '18 at 21:23
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Your proof is not complete. If your argument at the end is:-

Since there are two subsequences of an converging to two different limits, then, this implies that the sequence $(a_n)$ diverges.

Then, $(2)$ is redundant as it does not add anything to $(1)$ in the proof.

Consider the subsequence $\frac{(-1)^{2n}2n}{2n+1}$ of $(a_n)$. Then, $\lim\frac{(-1)^{2n}2n}{2n+1}=1$ as $n \to \infty \tag{1}$

Let $\epsilon>0$ be given. Choose $N\geq \frac{1}{\epsilon}$. Then, for all $n\geq N$, $|\frac{(-1)^{2n}2n}{2n+1}-1|<\epsilon \tag{2}$


An alternative method to do this which does not involve the subsequence notion is as follows:-

Assume that $\lim \frac{(-1)^n(n)}{n+1}=a$ as $n \rightarrow \infty$ for some $ a \in \mathbb{R}$.

Using definition of a limit, we get:

For every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that if $n > N$, then $\left|\frac{(-1)^n(n)}{n+1}-a \right| < \epsilon$.

In particular, let $\epsilon=\frac{1}{2}$.

Then, there exists $N \in \mathbb{N}$ such that if $n > N$, then $\left|\frac{(-1)^n(n)}{n+1}-a \right| < \frac{1}{2}$.

We have two cases to consider:-

$(a)$ Assume $n$ is odd:- If $n > N$, then $\left|\frac{-n}{n+1}-a \right| < \frac{1}{2}$.

$(b)$ Assume $n$ is even:- If $n > N$, then $\left|\frac{n}{n+1}-a \right| < \frac{1}{2}$.

Now, using triangle inequality, we get:

$\frac{2n}{n+1}=\left| \left(\frac{-n}{n+1}-a \right) - \left(\frac{n}{n+1}-a \right) \right| \leq \left| \left(\frac{-n}{n+1}-a \right) \right| + \left| \left(\frac{n}{n+1}-a \right) \right| < 1 $.

Note that $\frac{2n}{n+1}=2-\frac{2}{n+1}$ and so $\frac{1}{n+1} > \frac{1}{2}$.

This is equivalent to $n < 1$, a contradiction as $n \in \mathbb{N}$.

It follows that $(a_n)$ does not converge, hence $(a_n)$ diverges by definition.

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  • $\begingroup$ Thank you! However, how did you come up with the $\epsilon=1/2$? How did you know that choosing this particular value of epsilon will lead to a desired outcome? $\endgroup$ – A.Asad Mar 22 '18 at 20:08
  • $\begingroup$ Good question. If you see how I used the triangle inequality, I choose $C$ such that $\frac{2n}{n+1} < 2C$ leads to a contradiction. In fact, any number less than $\frac{1}{2}$ would do the same trick as well. $\endgroup$ – Dragonemperor42 Mar 22 '18 at 20:12
  • $\begingroup$ How did you go from 2-2/(n+1) to 1/(n+1)>1/2? Could you please elaborate a bit on that? $\endgroup$ – A.Asad Mar 22 '18 at 20:38
  • $\begingroup$ Sorry, I think I got it. for 2n/(n+1) to be less than 1, 1/(n+1) must be greater than 1/2. $\endgroup$ – A.Asad Mar 22 '18 at 20:41
  • $\begingroup$ Yes. I don't think I need to add this to my answer now. $\endgroup$ – Dragonemperor42 Mar 22 '18 at 21:24
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You're right, although I feel you should provide a little more justification for the inequalities you have writen. Also, consider this:

$$a_{2n}=\frac{(-1)^{2n}2n}{2n+1}=\frac{2n}{2n+1}=\frac{1}{1+\frac{1}{2n}}\to1$$ immediately, given that $\frac{1}{n}\to0$ - this is way much easier to prove, I think, from Archimedes-Eudoxus Principle.

Also, $$a_{2n-1}=\frac{(-1)^{2n-1}(2n-1)}{2n-1+1}=-\frac{2n-1}{2n}=-\frac{1-\frac{1}{2n}}{1}=-1+\frac{1}{2n}\to-1$$

So, since $a_{2n}$ and $a_{2n-1}$ converge to different numbers, $a_n$ is not convergent.

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    $\begingroup$ Χαιρετίζω την απάντησή σου συνάδελφε! $\endgroup$ – Michael Mar 22 '18 at 19:39
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    $\begingroup$ Ευχαριστώ πολύ, συνάδελφε! :) $\endgroup$ – Βασίλης Μάρκος Mar 22 '18 at 19:57
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    $\begingroup$ Please excuse me for being frank, but it is not very polite/respectful to use a language that most people using this site do not understand, even if it only states: "I welcome your reply, colleague!" respectively "Thank you, colleague! :)". $\endgroup$ – R.J. Etienne Mar 22 '18 at 20:29
  • $\begingroup$ You are absolutely right, my apologies! $\endgroup$ – Βασίλης Μάρκος Mar 22 '18 at 20:46
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Here is an alternative, although the difference is very superficial. Since

$$ a_n = (-1)^n\frac{n+1-1}{n+1} = (-1)^n - \frac{(-1)^n}{n+1} $$ you see that $(a_n)$ is the sum of a divergent sequence and a convergent sequence and therefore it must be divergent.

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For a convergent sequence all subsequences converge to the same limit (it is a theorem), thus it suffices to note that

  • $a_{2k}\to 1$
  • $a_{2k+1}\to -1$

to conclude that the given sequence does not concerge.

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