Convergence of an infinite sum Is it possible to use the comparison test for convergence in the following series?
$$\sum_{n=1}^\infty \sin \frac 1 n$$
The exercise says that I should find a linear function $f(x)$ that satisfies $f(x)<\sin(x)$ if $0<x<\pi/2$ but I don't understand how that would help.
Any ideas?
Thanks!
 A: Note that $g(x) = x/2$ satisfies $\sin(x) > g(x)$ for all $x < \pi/{2}$. Moreover, we have that
$$
\sum_{n=1}^\infty g(1/n) = \sum_{n=1}^\infty \frac{1}{2n}
$$
does not converge. What does this tell you?
A: Another approach (otra idea): since $\;\sin\frac1n>0\;\;\forall\,n\in\Bbb N\;$ , we can use the limit comparison test:
$$\frac{\sin\frac1n}{\frac1n}\xrightarrow[n\to\infty]{}1\implies \sum_{n=1}^\infty\sin\frac1n\;\;\text{converges}\;\;\iff\sum_{n=1}^\infty\frac1n\;\;\text{converges}$$
and since the rightmost one is the harmonic series....
A: We shall use the following:
Fact. If $x\in [0,1]$, then $\sin x\ge ax$, where $a=\cos( 1)>0$. 
Proof. For every $x\in [0,1]$
$$
\cos 1\le \cos x,
$$
and integrating in $[0,x]$ we obtain
$$
\int_0^x cos (1)\,dt\le \int_0^x\cos t\,dt,
$$
and hence
$$
ax= \big(\cos( 1)\big)\,x\le \sin x. \tag*{$\Box$}
$$
Now, for every $n\in\mathbb N$, using the above Fact we ontin 
$$
\sin\frac{1}{n}\ge \frac{a}{n},
$$
and as $\sum_{n=1}^\infty \frac{a}{n}$ diverges, so does $\sum_{n=1}^\infty\sin\frac{1}{n}$, by virtue of the comparison test.
A: $$
\sin\frac 1 n \geq \frac{1}{2n},
$$
so by comparison, the series your're asking about diverges.
The linear function involved is $x\mapsto x/2$.  That function is ${}<\sin x$ if $x$ is positive and sufficiently small.
