Convergence of the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$..? Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..
 A: In case you don't know the power series but do know that $\sin'(t) = \cos(t)$
and $\sin(t) \le t$ for $t \ge 0$:
$x-\sin(x) = \int_0^x (1-\cos(t))dt$.
$\cos(2t) = \cos^2(t)-\sin^2(t)
= 1-2\sin^2(t)$
so $1-\cos(t) = 2\sin^2(t/2) \le t^2/2$
since $|\sin(t)| \le |t|$.
Therefore
$x-\sin(x) \le \int_0^x (t^2/2)dt = x^3/6$
and the remainder of the proof goes through as before.
A: Yet you can show the convergence of $\sum_{n=1}^{\infty} \left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)$ as follows, which is based  just on the inequality \begin{gather*}\sin(x)<x,\qquad \forall x>0.\end{gather*}
Indeed, since for all $n\in\mathbb{N},$  $1/n>0,$ we arrive at $0<1/n-\sin(1/n),$ and so the series under consideration is of strictly positive terms. Now we show that \begin{gather*}\tag{$\star$}x-\sin(x)<x^2, \qquad \forall x\in (0,1].\end{gather*} To this end, introduce the auxiliary function $f: x\mapsto x^2-x+\sin(x).$ Then $f$ is certainly continuous on the interval $[0,1].$ Note that 
\begin{align*}
f'(x)&=2x-1+\cos(x)=2x-\big(1-\cos(x)\big)\\
&=2x-2\sin^2\left(\frac{x}{2}\right)\\
&>2x-2\cdot\left(\frac{x}{2}\right)^2\\
&=\frac{x}{2}(4-x)>0,\qquad \forall x\in(0,1),
\end{align*} 
and thus we conclude that $f$ is increasing strictly on $[0,1].$ As a result, 
$$x^2-x+\sin(x)=f(x)>f(0)=0,\qquad \forall x\in (0,1].$$  Hence $(\star)$ is valid.
Finally, by $(\star),$ we get 
\begin{gather*}
0<\frac{1}{n}-\sin\left(\frac{1}{n}\right)<\frac{1}{n^2}, \qquad \forall n\in\mathbb{N}.
\end{gather*}
Since $\sum_{n=1}^{\infty} 1/n^2$ converges,  we deduce that $\sum_{n=1}^{\infty} (1/n-\sin(1/n))$ is convergent, by comparison test. 
Furthermore, if you feel the above argument is too clumsy, try as follows. It is easy to show that (for instance, by L'Hospital's Rule) 
\begin{gather*}\lim_{x\to 0} \frac{x-\sin(x)}{x^2}=0<1,\end{gather*}
and then, by the locally order-preserving property of limit, there exists some $\delta>0$ such that 
\begin{gather*}
x-\sin(x)<x^2,\qquad \forall x\in (0, \delta).\end{gather*}
Therefore, there exists some natural number $N$ such that $1/N<\delta.$ Hence 
\begin{gather*}
\frac{1}{n}-\sin\frac{1}{n}<\frac{1}{n^2}, \qquad \forall n>N.\end{gather*}
Then the desired result follows.
A: Consider that 
$$\frac{1}{n} - \sin{\left(\frac{1}{n}\right)} \sim \frac{1}{6 n^3}$$
as $n \rightarrow \infty$.  Then use the comparison test.
A: We know the fact that for any natural number n , there holds an well- known in-equality that $1/(n+1) < \operatorname{sin}(1/n) <=1/n$ , then,   $(-\operatorname{sin}(1/n)) < -1/(n+1)$ , for all $n \in \mathbb N$ ;
$0 < (1/n - \operatorname{sin}(1/n)) < (1/n - 1/(n+1)) <1/(n(n+1)) 
<1/n^{2 }$,  hence as :-  by the comparison test, the result follows .
