<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] Here is another approach:
show that there exists infinitely many terms within $\delta$ neighbourhood of 0,1 and -1.
in other words 0,1 and -1 are all limits so there is no single limit for this sequence.
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