Prove divergence of sequence $a_n=1+n\sin\left(\frac{n\pi}{2}\right)$

So the above sequence is obviously not convergent because of the sine, but how could one prove this? Thanks for any pointers.

Tom
Sequence $a_n=1+n\sin\left(\frac{n\pi}{2}\right)$ has a subsequence 
$$b_0=a_1,\ b_1=a_5,\ b_2=a_9,\ \ldots,\ b_k=a_{4k+1},\ \ldots$$
of a form $b_k = 4k+2$ which is divergent. Hence sequence $\{a_n\}$ is divergent.

UPD:
$$b_k = a_{4k+1} = 1 + (4k+1)\sin\left(\frac{(4k+1)\pi}{2}\right)= 
1 + (4k+1)\sin\left(2k\pi + \frac{\pi}{2}\right) = 
$$
$$
= 1 + (4k+1)\sin\left(\frac{\pi}{2}\right) = 1 + 4k+1 = 4k+2
$$
We know that a sequence if diverging to $\infty$ iff $\forall G \in \mathbb{R}$, no matter how large it is, $\exists n \in \mathbb{N}$ such that $a_k&gt;G \; \forall k \geq n$. Well, we can easily find any odd number $&gt;m$ and then the property will hold for some of the terms (all the terms $a_k, k&gt;G, k \equiv 1 \; (\mathrm{mod}\;2))$. So, we have a diverging subsequence. But, we have another subsequence $\{1,1,1,...\}$ which converges to $1$. So, the sequence is neither converging nor diverging to $\infty$, it's infinitely oscillating.

BUT, if we consider divergence as the negation of convergence, it's definitely diverging as it has a diverging subsequence.