Question

Show that the series 1 - cos(1/n) converges absolutely ?

Answer 1

let it be a function f(n)=1-cos(1/n)

as 'n' increases and approach infinity, (1/∞) will become 0. cos(0)=1
therefore, 1-cos(1/∞)=1-cos(0)=1-1=0 the function converges to 0.

Answer 2

\[\lim_{n \rightarrow \infty}(1-\cos (\frac{ 1 }{ n }))\]\[= (1-\cos (\frac{ 1 }{ \infty }))\]\[= (1-\cos (0))\]\[= (1-1)\]\[= 0\]

Answer 3

@NicksonYap
Yes, as a sequence, it is convergent, but what about as a series? :)
\[\Large \sum_{n=1}^\infty \left[1-\cos\left(\frac1n\right)\right]\]

Answer 4

ah, don't really know how to do this.

The series is convergent because as 'n' increases, cos(1/n) gets closer and closer to 1, making 1-cos(1/n) closer and closer to 0.
So the series will end at some point.

Can we use integration instead?

Answer 5

You can certainly try to integrate it. If you can prove that the integral is convergent, then the summation will also be convergent.

LOL not all sequences that converge to zero are convergent as a series. Harmonic series, anyone? :)
\[\Large \sum_{n=1}^\infty \frac1n \qquad \qquad \color{red}{\text{DIVERGENT!!!}}\]

Answer 6

Why do we start from n=1? why not n=0?

Answer 7

We simply can't start with n = 0?
lol
You try evaluating \[\Large 1 - \cos\left(\frac1n\right)\]
when n = 0
I'll wait ;)

Answer 8

You are right, the series can only start with n = 1 for its denominator is n. Anyway, is there anyway to change this series into a simple function which is possible to integrate?

Answer 9

No idea. However...

Answer 10

Are you panicking? :D

Answer 11

My friend came up with the idea that \[1 - \cos \frac{ 1 }{ n } = \frac{ 1 - \cos(u) }{ 1 - u} \times (1-u) = (1 - u) sinu\]

Answer 12

but i don't get it :))

Answer 13

Because it doesn't make sense? lol

Answer 14

it doesn't make sense to me because i don't know which equation is this. Do you know this or is it available ?

Answer 15

Pay no heed to it.
Use the Limit comparison test ^_^

Answer 16

try expanding cos(1/n)

Answer 17

@experimentX don't we have enough summations as it is? :D
jk

Answer 18

no no ... just approximate it.
1 - cos(1/n) < 1/n^2

Answer 19

^ A bit hard to prove, methinks.
LCT with 1/n^2 works ^_^

Answer 20

but how can you come up with 1/n^2 ?

Answer 21

you get that after expanding cos(1/n)

Answer 22

oh... okay :(

Answer 23

@sukanguyen
To use the LCT, you have to have a suspicion of sorts, whether or not it's convergent, so suppose it's convergent....

We'll find a *known* convergent series with which to compare limits.

We know the series \[\Large \sum_{n=1}^\infty \frac1{n^2}\] is convergent, so get this limit:
\[\LARGE \lim_{x\rightarrow \infty}\frac{1 - \cos\left(\frac1x\right)}{\frac1{x^2}}\]

If it goes to some number, then that will be enough to prove that your series is convergent ^_^

Answer 24

well well ... that works too.

Answer 25

Of course it does... I thought long and hard about it >:(

^_^

Answer 26

if we can't start with n=0, why choose n=1? why not start with like n=0.00000001? It was not said that 'n' is discrete..

Answer 27

It's a series, nick.
A series is a sequence of partial sums.
The domain of a sequence is the set of natural numbers.

^_^

Answer 28

it's pointless to involve singularities due to zero.

Answer 29

While we're at it @sukanguyen
Try to evaluate that limit I gave you ^_^

Answer 30

It's pretty hard :( as i prove,the limit approaches infinity

Answer 31

No it isn't. In fact, it's indeterminate, heard of indeterminate forms? ^_^

Answer 32

use L'hopital ... if you are doing sequence and series then I believe you have done L'hopital
http://www.wolframalpha.com/input/?i=Limit%5Bx%5E2+%281++-+Cos%5B1%2Fx%5D%29%2C+x+-%3E+Infinity%5D
although there are other ways you can do it.

Answer 33

easy way is to expand cos(1/x) using Taylor series.
the other way it to write 1 as cos(0)
and use cos(C)-cos(D) formula .... you should end up with the same value.

Answer 34

i know it's inderterminate. L'hospital is possible also

Answer 35

Okay, i get it. Thank you guys :)