How many limit points does $S_n = \sin^2(\pi n/2)+(-1)^n/{\pi n}$ have?
I think it has only one limit point, but I'm not so sure since I'm confused by the definition of limit points I see online (they use topology?). I'd appreciate if someone could show me how to do this.
It seems to me since $\sin^2(\pi n/2) = 1$ that we $\displaystyle S_n = 1+\frac{(-1)^n}{\pi n}.$
So $\displaystyle S_{2n} = 1+\frac{1}{2\pi n} \to 1$ and $\displaystyle S_{2n+1} = 1-\frac{1}{\pi(2n+1)} \to 1$
So no subsequence of $S_n$ converges anything other than $1$?
Consider the case when $n=2k$ for $k \in \mathbb{N}$. Observe that we have
$$S_{2k} = (\sin(k\pi))^{2}+\frac{1}{2\pi k}= \frac{1}{2\pi k}$$. Hence the limit to infinity is zero.
Clearly the subsequence $S_{2k}$ and $S_{2k+1}$ do not have the same limit value as $k$ tends to infinity.