Prove that function $f(n) = \cos(1/n)$ is monotonically decreasing? Ho can we prove that this function
$$f(n) = \cos(1/n)$$
is monotonically decreasing? I computed derivative of this function and got:
$$f'(n) = \frac{\sin(1/n)}{n^2}$$
and it is positive, but should be negative, because function $f(n) = \cos(1/n)$ is monotonically decreasing.  
 A: Is $n$ supposed to be a positive integer? If so you should probably call $f$ a sequence instead of a function, to prevent confusion.
Assuming that yes you are talking about the sequence $\cos(1/n)$: That sequence is not decreasing, it's increasing.
In detail: The sequence $1/n$ is decreasing. And $0<1/n<\pi$, which means that $\cos$ is also decreasing on the relevant range. A decreasing function of a decreasing sequence is increasing.
A: Ignoring the derivative which does show $f(x) = \cos(1/x)$ is increasing when $\sin (1/x)/x^2$ is positive (which is the case for all $x > 1/\pi$ [which is the case for all natural numbers n]), "common sense" should indicate $f(x) = \cos(1/x)$ is increasing for significantly large $x$.
As $x \rightarrow \infty$ then $1/x$ decreases to 0 (as a limit, of course).  As $v = 1/x$ decreases from $v = \pi$ ($x = 1/\pi$) to 0, $\cos v$ increases from -1 to 1 (as a limit, of course). 
So $f(x)$ is monotonically increasing on $[1/\pi, \infty)$.  
But $f(x)= \cos(1/x)$ is monotonically decreasing on $(-\infty, - 1/\pi]$. (But then it oscilates innumerably, is undefined at x = 0, oscilates innumerable times between 0 and $1/\pi$ and then increases in value from -1 to a limit of 1.)
A: The function $x\mapsto\cos(1/x)$ is not decreasing in $(0,\infty)$, because its derivative is
$$
\frac{\sin(1/x)}{x^2}
$$
and obviously this assumes positive and negative values over suitable intervals.
The sequence $n\mapsto\cos(1/n)$ cannot be decreasing, because
$$
\lim_{n\to\infty}\cos\frac{1}{n}=1
$$
and $\cos(1/n)<1$ for all positive integers $n$.
Let's see if it is increasing, that is,
$$
\cos\frac{1}{n}<\cos\frac{1}{n+1}
$$
which is true, because $1/n\in(0,\pi/2)$ for all positive integers $n$ and the cosine function is decreasing in the interval $(0,\pi/2)$.
A: Remember that the derivative of $1/n^2$ is negative.
