<s>[INST] Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor oscillates between -1, 0 and 1.
Here is my attempt to prove that it is divergent:
Suppose that the sequence is convergent. This should lead to a contradiction. If it is convergent, then, for every $\epsilon>0$, there is an $N>0$ such that if $n>N$, then $\left|\dfrac{n}{n+1}\sin\dfrac{n\pi}{2} -L\right|<\epsilon$.
In particular, for $\epsilon = \frac{1}{2}$:
$$-\frac{1}{2}<\frac{n}{n+1}\sin\frac{n\pi}{2} -L<\frac{1}{2} \text{ for every integer }n>N$$
Notice that the sequence of values of $\sin\dfrac{n\pi}{2}$ is $1, 0, -1, 0, 1, 0, -1, \cdots$ for $n=1, 2, 3, 4, 5, 6, 7, \cdots$. If $n = 1, 5, 9 \cdots$, $\sin\frac{n\pi}{2}$ is $1$, so, if we choose a particular $n$ from this list of values, the value of $a_n$ is $\dfrac{n}{n+1}$. In this case, $\sin\dfrac{(n+2)\pi}{2}$ will be $-1$, therefore $a_{n+2} = -\dfrac{n+2}{n+3}$. So, we have that:
$$-\dfrac{1}{2}<\dfrac{n}{n+1}-L<\dfrac{1}{2} \text{ and } -\dfrac{1}{2}<-\dfrac{n+2}{n+3}-L<\dfrac{1}{2}$$
Rearranging the terms:
$$\dfrac{1}{2}>-\dfrac{n}{n+1}+L>-\dfrac{1}{2} \text{ and } \dfrac{1}{2}>\dfrac{n+2}{n+3}+L>-\dfrac{1}{2}$$
$$\dfrac{n}{n+1}+\dfrac{1}{2}>L>\dfrac{n}{n+1}-\dfrac{1}{2} \text{ and } \dfrac{1}{2}-\dfrac{n+2}{n+3}>L>-\dfrac{n+2}{n+3}-\dfrac{1}{2}$$
The first inequality says that $L$ is positive, but the second inequality says that $L$ is negative; if this is correct so far, then I found a contradiction, proving that the sequence is divergent.
My question is: is this correct or is there some inconsistency?
 [/INST] Claim for you to prove: if a sequence converges then any subsequence converges and to the very same limit as the whole sequence.
Well, what about choosing two very special subsequences here and proving they don't have the same limit? In fact, one of the "obvious" subsequences doesn't even have a limit...
</s>