Why is cos(1/n) an increasing function?

[phyzz]

Homework Statement

I have the series an = (-1)^{n} cos(1/n) and I have to determine whether it converges or diverges.

Homework Equations

I used the Leibniz criterion

The Attempt at a Solution

However, I determined that bn = cos(1/n) is a decreasing function because:

n+1 > n

1/(n+1) < 1/n

cos(1/(n+1)) < cos(1/n)

ie cos(1/(n+1)) - cos(1/n) < 0

so bn+1 - bn is < 0 meaning it's decreasing no?

I said it diverged because lim n -> ∞ bn = 1 (and not 0 which is the condition for it to converge)

I got the answer correct, but for the wrong reasons...

How do you show that cos(1/n) is an increasing function? Or any function for that matter? I thought my n+1 > n method was valid most (if not all) functions. Obviously not :(

Thanks a lot!

 

[Mentallic]

1/(n+1) < 1/n

cos(1/(n+1)) < cos(1/n)

Is this true? Think about the value of cos(x) for x approaching zero.

 

[phyzz]

Is this true? Think about the value of cos(x) for x approaching zero.

I don't know whether it's true or not, that's the thing
cos(x) x approaching 0 is 1 right?

 

[phyzz]

[cos(1/n)]' = - sin(1/n) / n^2

ie f'(n) < 0 for all n because of the negative sign, so it should be decreasing?

 

[phyzz]

ok I understood

n+1 > n

1/(n+1) < 1/n

sin(1/(n+1)) < sin(1/n)

ie sin(1/(n+1)) - sin(1/n) < 0

BUT if it's cos:

n+1 > n

1/(n+1) < 1/n

cos(1/(n+1)) > cos(1/n)

ie cos(1/(n+1)) - cos(1/n) > 0

gonna learn it like:

cos of something with a larger denominator is > cos of something with smaller denominator and
sin of something with a larger denominator is < sin of something with smaller denominator

 

[Mentallic]

Yes, cos(x) is decreasing for 0<x<pi/2 which means that if we take a positive integer n, then 1/n > 1/(n+1) and as such, because cos(x) is decreasing, cos(1/n) < cos(1/(n+1))

 

[phyzz]

Thanks!

 

[Mentallic]

You're welcome! And by the way,

I don't know whether it's true or not, that's the thing
cos(x) x approaching 0 is 1 right?

What I wanted you to think about is that as x approaches 0 (thus getting smaller and smaller) then the value of cos(x) approaches 1 (thus getting bigger and bigger).

 

[Mute]

1/(n+1) < 1/n

cos(1/(n+1)) < cos(1/n)

Just in case it's not entirely clear to you, when

x < y,

it follows

f(x) < f(y)

only if x and y are both contained in a range for which f(x) is a monotonically increasing function: f'(x) > 0.

If the function is monotonically decreasing (f'(x) < 0) on a range for which both x and y are in, then

f(x) > f(y).

This second case corresponds to the problem here.

Lastly, note that these are not the only possibilities. If f(x) is not monotonic on a range that contains both x and y, you can get any of f(x) < f(y), f(x) > f(y) or f(x) = f(y) for x < y. For example, if ##y = x + 2\pi##, then x < y, but for f(x) = cos(x), f(x) = f(y).

 

[scurty]

[cos(1/n)]' = - sin(1/n) / n^2

ie f'(n) < 0 for all n because of the negative sign, so it should be decreasing?

##\frac{d}{dn} \cos{(\frac{1}{n})} = - \sin{(\frac{1}{n})} \cdot (- \frac{1}{n^2})##

You missed a minus sign.