Proof Verification: $\frac{(-1)^{n}n}{n+1}$ diverges 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?
 A: 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.
A: 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.
A: 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.
A: 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.
