Proving $a_n = \frac{(-1)^n n}{n+1}$ has no limit

Question

I want to prove the sequence $$a_n = \frac{(-1)^n\cdot n}{n+1}$$ has no limit as $\,n\to+\infty$.

I tried to work with inequalities to get that $\,|a_n-L|\geqslant\varepsilon\,$ and the way I tried to solve it was to divide into cases:
$\,L\geqslant0\,$ or $\,L<0\,.$
When I did the case with $\,L\geqslant0\,,\,$ I chose an even $\,n\,$ and I found there is $\,\varepsilon=L\,$ for which this is true, but I don't know if I am allowed to represent $\,\varepsilon\,$ with $\,L\,$ or should I represent it as a number. And I would also be glad if someone could show me the way to solve this because my prof hasn't shown us this although he gave us to do that.

Answer 1

This is how to do it using the definition of limits.

We say that $a_n$ converges (to $L$) iff $$\exists L \in \mathbb{R} \; \forall \varepsilon > 0 \; \exists n_0\in\mathbb{N} \; \forall n\geq n_0 : |a_n - L| < \varepsilon.$$ Hence, we say that $a_n$ does not converge iff $$\forall L\in\mathbb{R}\; \exists \varepsilon > 0 \; \forall n_0\in\mathbb{N} \; \exists n\geq n_0 : |a_n - L| \geq \varepsilon,$$ or in other words, you want to find an $\varepsilon > 0$ such that $|a_n - L| \geq \varepsilon$ is satisfied for infinitely many $n$.

Let $L\in\mathbb{R}$ be given and take $\varepsilon = |L|$ (or, if $L=0$, set $\varepsilon = 1/2$). Let moreover $n_0\in\mathbb{N}$ be given.

Case 1: $L > 0$. Define $n = 2n_0 + 1 > n_0$. Then $(-1)^n = -1$ and $$|a_n - L| = L -\frac{-(2n_0 + 1)}{2n_0 + 2} = L + \frac{2n_0 + 1}{2n_0 + 2} > L = \varepsilon.$$ Case 2: $L < 0$. Define $n = 2n_0 > n_0$. Then $(-1)^n = 1$ and $$|a_n - L| = \frac{2n_0}{2n_0 + 1} - L > -L = \varepsilon.$$ Edit: Case 3: $L=0$. Define $n=n_0+1$. Then $$|a_n - L| = \frac{n}{n+1} \geq \frac{1}{2} = \varepsilon,$$ where $\frac{n}{n+1} \geq \frac{1}{2}$ for $n\geq 1$ is an easily verified inequality.

Since this covers both cases, we are done.

Answer 2

Alternative approach:

When dealing with oscillating sequences, the triangle inequality is handy.

That is, for any $~a_r,a_s,L \in \Bbb{R},~$

$$|a_r - a_s| \leq |a_r - L| + |L - a_s|. \tag1 $$

Therefore, when constructing an $~\epsilon,N~$ proof, if you can specify a specific value for $~\epsilon~$ such that no matter how far in the sequence that you go, you will always be able to find $~a_r, a_s,~$ beyond that point in the sequence such that $~|a_r - a_s| > 2\epsilon,~$ then you are done.

The reason is that no matter what $~L~$ is chosen, it will have to obey the inequality in (1) above. So, if the LHS of (1) above is $~> 2\epsilon,~$ it will be impossible for both terms on the RHS of (1) above to be $~\leq \epsilon.$

In the present case, with $~\displaystyle \lim_{n\to\infty} |a_n| = \lim_{n\to\infty} \dfrac{n}{n+1} = 1,~$
and with the elements $~a_n~$ oscillating between positive and negative,
simply specify that (for example) $~\epsilon = \dfrac{1}{2}.$

If you can show that when $~\epsilon = \dfrac{1}{2},~$ that it is impossible for there to exist any value $~L,~$ such that beyond a certain point $~a_N~$ in the sequence, all of the following elements satisfy $~|a_n - L| \leq \epsilon,~$ then you are done.

Using the idea in (1) above, the problem reduces to showing that for any $~N \in \Bbb{Z^+},~$ there will always exist $~r,s > N,~$ such that $~|a_r - a_s| > 2\epsilon = 2 \times \dfrac{1}{2} = 1.$

Given the oscillating nature of the sequence, and given any value $~N,~$ choose any $~r,s > N,~$ such that $~r~$ is odd and $~s~$ is even.

Then, you will have that

$$a_r < \frac{-1}{2}, ~a_s > \frac{1}{2} \implies |a_r - a_s | > 1 = 2\epsilon.$$

Answer 3

The easiest way to prove this is to show that subsequence of even indices converges to 1 ($x_{2n}=\frac{2n}{2n+1}$) and subsequence of odd indices converges to −1 ($x_{2n-1}=-\frac{2n-1}{2n}$). This is in contrary with the theorem: If a sequence has a limit x, then all subsequences of that sequence have the same limit x.

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EDIT: It can be easily derived from the previously said that there exists a limit of a sequence $y_n=(-1)^nx_n$ iff $\lim_{n \to \infty}x_n=0$. It is also clear that $\lim_{n \to \infty}y_n=0$

Answer 4

The $(-1)^n$ in the numerator indicates that $a_n$ is positive if $n$ is even and negative if $n$ is odd. Therefore, there is only one number this sequence may converge to. It suffices to show that $\frac{n}{n+1}$ does or does not converge to that number.

Answer 5

$$\left|a_{n}-a_{n-1}\right|=\left|(-1)^n \left( \frac {n(n+2)+(n-1)(n+1)}{n(n+1)} \right)\right|$$ $$=\frac{2n^2+2n-1}{n(n+1)}$$ $$2-\frac 1 {n(n+1)}$$

If $n>0$, we are subtracting at most $\frac 1 2$, so the difference is greater than $1$.

Therefore, this is not a Cauchy sequence, and thus does not converge

Answer 6

Assume $c_n\to c\neq 0.$ Then the sequence $a_n=b_nc_n$ is convergent iff $b_n$ is convergent, as $b_n=a_n/c_n.$ Apply this to $c_n=n/(n+1)\to 1$ and $b_n=(-1)^n.$ Since $b_n$ is divergent, so is $a_n.$

Another approach: assume the sequence $b_n-a_n$ is convergent.Then $a_n$ is divergent iff $b_n$ is divergent. We can apply that to $a_n=(-1)^n{n\over n+1}$ and $b_n=(-1)^n,$ as $b_n-a_n=(-1)^n{1\over n+1}\to 0.$

Answer 7

Here's another way to do it using neighborhoods and the fact that for all convergent sequences $\vec{p}$ with limit $x$, it holds for all neighborhoods $U$ of $x$, that finitely many elements fall outside of $U$.

For the purposes of this sequence, I'll assume that $a$ starts at $a_0$ which is $0$.

I'll show that $\vec{a}$ can't approach anything by pointing out explicit neighborhoods of every real number with infinitely many points outside of them.

I claim that $\vec{a}$ does not approach $0$. I choose as my neighborhood the open interval $(-\frac{1}{10}, \frac{1}{10})$. Aside from $a_0$, all other elements of the sequence are outside of this set.

Let $w$ be a positive number. There are infinitely many elements of $\vec{a}$ outside of $(w/2, 2w)$ because the elements associated with positive odd indices are negative.

Let $w$ be a negative number. There are likewise infinitely many elements of $\vec{a}$ outside of $(2w, w/2)$ because the elements associated with positive even indices are positive.

Thus $\vec{a}$ does not converge.