Show that a sequence does not have a limit

Question

How can I show that the sequence $x_n=\frac{(-1)^nn}{n+1}$ does not have a limit by using only the definition of limits?

Attempts:

Let's assume that our sequence has a limit $x$. Then there exists such a number $N$ so that for any $n>N$ $|a_n-a| < \epsilon$. This is also true for $\epsilon = \frac{1}{2}$. Now I take the numbers $N_1$, $2N_1$ and $2N_1+1$. But when I start subtracting (just like it is done with the sequence $x_n=(-1)^n$), I get a result which has $N_1$ in it, and thus I can not get a contradiction. What do I do now?

Answer 1

Intuition: The problem with this sequence is that it really should have two limits: $1$ and $-1$. (Consider separately what happens for odd $n$ and even $n$.)

Sketch of proof: To leverage our intuition into rigor, one way to proceed would be consider what the limit might be. If the sequence converges to $1$, then we have a problem because for odd $n$ the sequential values are negative and in particular at least one unit away from $1$. On the other hand, if the sequence converges to some number (call it $L$) that isn't $1$, then $L$ is separated from $1$ by some positive distance; this will be a problem when we consider the even subsequence, which will get arbitrarily close to $1$.

Answer 2

HINT

Let consider the following subsequences

• $n=2k\to \infty \implies x_{2k}=\frac{(-1)^{2k}2k}{2k+1}\to \,?$

• $n=2k+1\to \infty \implies x_{2k+1}=\frac{(-1)^{2k+1}2k+1}{2k+3}\to \,?$

and recall that if a sequence tends to $L$ then all the susequences must tends to the same limit $L$.

Answer 3

If $x_n=\frac{(-1)^nn}{n+1} $ has a limit $v$, then, for any $c > 0$ there is a $n(c)$ such that $|x_n-v| < c$ for all $n > n(c)$.

In this case, $x_n$ has two subsequences that have limits: $x_{2n} \to 1$ and $x_{2n+1} \to -1$.

So let's choose a $c$ that is less than half the distance between the limits of these two subsequences, for example, $c = \frac12$.

Then for $n > n(c)$ $|x_n-v| < c$. Suppose also that $n > 10$. Then $x_{2n} = \dfrac{2n}{2n+1} =1- \dfrac{1}{2n+1} $ so $c \gt |x_{2n}-v| = |1- \dfrac{1}{2n+1}-v| \ge |1-v|- \dfrac{1}{2n+1} $. Therefore $|1-v| \le c+\dfrac{1}{2n+1} $ or $-c-\dfrac{1}{2n+1} \le 1-v \le c+\dfrac{1}{2n+1} $ or $1-c-\dfrac{1}{2n+1} \le v \le 1+ c+\dfrac{1}{2n+1} $.

Similarly, $x_{2n-1} = -\dfrac{2n-1}{2n} =-1+\dfrac{1}{2n} $ so $c \gt |x_{2n}-v| = |-1+ \dfrac{1}{2n}-v| \ge |-1-v|- \dfrac{1}{2n} $. Therefore $|-1-v| \le c+\dfrac{1}{2n} $ or $-c-\dfrac{1}{2n} \le 1-v \le c+\dfrac{1}{2n} $ or $-1-c-\dfrac{1}{2n} \le v \le -1+ c+\dfrac{1}{2n} $.

Therefore $1-c-\dfrac{1}{2n+1} \le v -1+ c+\dfrac{1}{2n} $ so $2 \le 2c+\dfrac{1}{2n}+\dfrac{1}{2n+1} $ which is false for $c \le \frac12$ and $n > 10$.

This is readily modified to show that any sequence with two subsequences that have different limits can not iself have a limit. Even more general results can be similarly proved.