Show that a sequence does not have a limit 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?
 A: 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 $1$.
A: 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 $L$.
A: 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.
