Contradictory results about the convergence of $\sum_{n=1}^{\infty} \sin(2\pi n)$ By observation, I can conclude that $$\sum_{n=1}^{\infty} \sin(2\pi n)=\sin2\pi+\sin4\pi+\dots=0+0+\dots=0$$
using the fact that $\sin(2\pi n)=0$ for all $n\in\mathbb{N}$.
However, I also convinced myself that $$\lim_{n\to\infty}\sin(2\pi n)\neq0$$
by the following argument:
Suppose $\epsilon<1$ and there exists an $N>0$ such that $n>N$ implies $\lvert\sin(2\pi n)-0\rvert=\lvert\sin(2\pi n)\rvert<\epsilon<1$.
Since the period of $\sin(2\pi n)$ is $1$, then there exists some $x\in(N, N+1]$ such that $\lvert\sin(2\pi x)\rvert=1$ is not less than $\epsilon$.
The limit therefore cannot be $0$.
By the $n$-th term divergence test, this implies that my original series diverges.
Why are my conclusions contradictory? I tried Wolfram Alpha and the engine seems to suggest that the series both converges and diverges.
 A: There's an "implicit domain" issue going on. There are actually two functions at play here:

*

*$f: \color{red}{\mathbb{R}}\rightarrow\mathbb{R}: x\mapsto \sin(2\pi x)$


*$g: \color{red}{\mathbb{N}}\rightarrow\mathbb{R}:x\mapsto\sin(2\pi x)$.
These two functions have different domains (highlightedd in red), and this is crucial. When each claim is appropriately interpreted, $$\lim_{x\rightarrow\infty}f(x)=0$$ is false but $$\lim_{x\rightarrow\infty}g(x)=0$$ is true. For the $n$th term divergence test, it is the latter claim that is actually relevant, and we can see that this should be the case since when asking about $\sum_{i\in\mathbb{N}}f(i)$ it's obvious that only the behavior of $f$ on natural number inputs is relevant.
It may help at this point to state the relevant divergence test more clearly (to get the more familiar version of the test, take the contrapositive):

Suppose $h:\mathbb{R}\rightarrow\mathbb{R}$ is a function such that $\sum_{i\in\mathbb{N}}h(i)$ converges. Then for every $\epsilon>0$ there is some $N$ such that for every integer $n>N$ we have $\vert h(n)\vert<\epsilon$.

It should be clear, due to the explicitly bolded part (normally only implicit in the notation - specifically, the use of "$n$" as a free variable) that there is in fact no paradox of the type you are concerned about.
