Prove that function $f(n) = \cos(1/n)$ is monotonically decreasing?

Question

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.

Answer 1

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.

Answer 2

Remember that the derivative of $1/n^2$ is negative.

Answer 3

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 $(0,\pi/2)$.

Answer 4

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.)