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

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.

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.