Analyze the convergence or divergence of the sequence $\left\{\frac1n+\sin\frac{n\pi}{2}\right\}$ 
Analyze the convergence or divergence of the following sequence

a) $\left\{\frac{1}{n}+\sin\frac{n\pi}{2}\right\}$
The first one is divergent because of the in $\sin\frac{n\pi}{2}$ term, which takes the values, for $n = 1, 2, 3, 4, 5, \dots$:
$$1, 0, -1, 0, 1, 0, -1, 0, 1, \dots$$
As you can see, it's divergent. To formally prove it,I could simply notice that it has constant subsequences of $1$s, $0$s, and $-1$s, all of which converge to different limits. If it were as subsequence, they would all be the same limit.
My procedure is that correct?

You could also use the fact that if a sequence is convergent, then the difference between successive terms must converge to $0$.
Let $x_n = \frac{1}{n}+\sin \frac{n \pi}{2}$. Then $x_{n+1}-x_n = \cos \frac{n \pi}{2} - \sin \frac{n \pi}{2} -\frac{1}{n(n+1)}$, from which we get the estimate $|x_{n+1}-x_n| \geq |\cos \frac{n \pi}{2} - \sin \frac{n \pi}{2}| -\frac{1}{n(n+1)} = 1-\frac{1}{n(n+1)}$. So the difference does not converge to $0$, so you can conclude that the sequence is divergent.
