If the series $\sum_i \sin x_i$ diverges does $\sum_i x_i$ necessarily diverge?
I feel like this should be a simple series comparison test or something, but I cannot figure a proof out. If you could point me to a duplicate that would also be helpful (can't find anything...) Thanks.
Edit: If it makes it easier, we can suppose that $x_i > 0$.
$\sin(x)$ $\le$ $x$:
Theorem: If $f$ is a continuous function on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ then there exists a number $c∈(a,b)$ such that $f′(c)=\frac{f(b)−f(a)}{b−a}$.
C1: Suppose that $x=0$. Then, $0=\sin(0)$ $\le$ $x=0$
C2: Suppose that $0 \lt x\lt 1$. Let $f(t)=\sin(t)$. Then $f$ is continuous and differentiable everywhere. In particular, $f$ is continuous on $[0,x]$ and differentiable on $(0,x)$. By the Mean Value theorem there exists number $c∈(0,1)$ such that:
$f′(c)=\frac{f(x)−f(0)}{x−0}$
The derivative of $\sin x$ is $\cos x$. Therefore:
$\cos(c)=\frac{\sin(x)−\sin(0)}x$
$\cos(c)= \frac{\sin(x)}{x}$
Note that the cosine function is bounded, that is, $−1≤cos(t)≤1$ for every real number t. Therefore: $−1≤\frac{\sin x}x≤1⇒\frac{\sin x}x≤1$
Since $0<x<1$, we multiply both sides of the inequality above to get that $sinx≤x$.
C3: Suppose that $1≤x<∞$. We know that $sin(t)$ is a bounded function and $−1≤\sin(t)≤1$ for every real number $t$. Thus $\sin x≤1≤x$, i.e., $\sin x≤x$.
As such, $sin(x)$ $\le$ $x$ for any $x∈R$.
Then,
$\sum_{i} \sin x_i$ ≤ $\sum_{i} x_i$, because any element of the first sum will be smaller or equal to the corresponding element of the second sum, because $\sin(x)$ $\le$ $x$ for any $x∈R$.
If $\sum_{i} \sin x_i$ diverges and $\sum_{i} \sin x_i$ ≤ $\sum_{i} x_i$, that means: $\sum_{i} x_i$ must diverge as well.