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’re on the right track, but you’ve left out an important step: you haven’t said anything to take the $1/n$ term into account. It’s obvious what’s happening, but you still have to say something.
Let $a_n=\frac1n+\sin\frac{n\pi}2$. If $\langle a_n:n\in\Bbb Z^+\rangle$ converged, say to $L$, then the sequence $\left\langle a_n-\frac1n:n\in\Bbb Z^+\right\rangle$ would converge to $L-0=L$, because $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ converges to $0$. Now make your (correct) argument about $\left\langle\sin\frac{n\pi}2:n\in\Bbb Z^+\right\rangle$ not converging and thereby get a contradiction. Then you can conclude that $\langle a_n:n\in\Bbb Z^+\rangle$ does not converge.
