Question

Sigma n=1 to infinity of cos n*pi/2 ??

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

so:
\[\sum_{n=1}^{\infty}\frac{\cos(n \pi)}{2} ?\]

Answer 3

\[\sum_{1}^{infinity} \cos(nPi/2)\]

Answer 4

yes!

Answer 5

What do you want to know? If it converges?

Answer 6

yes..if it converges and its sum if it converges

Answer 7

H/o just to make sure I'm rite is it cos(npi/2) or cos(npi)/2? Sorry, I just want to make sure

Answer 8

first one cos(npi/2)

Answer 9

> sum((1/2)*cos*n*Pi, n = 1 .. infinity);
"(->)"

infinity = 0

Answer 10

@inik: doesn't it depend on n??

Answer 11

Actually, it does not converge.

Answer 12

yes , i second male violence

Answer 13

It doesn't converge. Because if you take the lim n->infinity you get D.N.E. For a sum to converge, the lim n->infinity has to be ZERO.

Answer 14

if n is odd even doesn't it converge?

Answer 15

If n is odd it does. To zero. But if it is even it switches between 1 and -1.

Answer 16

so what must be the final answer..infinity and it diverges?

Answer 17

Not necessarily infinity. But it just diverges. You can diverge without being infinite.

Answer 18

Thanks :)

Answer 19

No problem :P

Answer 20

\[\sum_{n=1}^{\infty}(\cos n \Pi)/2^{n}\]

Answer 21

converges or diverges? if converges what is the sum?

Answer 22

Okay. So, we agree, that for no matter what value of n you plug in. cos(npi) is either +1 or -1? Starting with n=1 you get cos(pi)=-1 so:
\[\sum_{n=1}^{\infty}\frac{\cos(n \pi)}{2^n}=\sum_{n=1}^{\infty}\frac{(-1)^n}{2^n}\].
Well, that is the same as:
\[\sum_{n=1}^{\infty}(\frac{-1}{2})^{n}\].
This is ALMOST a geometric series. You notice if you do n=1 the series DOES NOT start at zero. So reindex it:
\[\sum_{n=1}^{\infty}(\frac{-1}{2})(\frac{-1}{2})^{n-1}\].
So you know a geometric series converges if the absolute value of the radius is less than 1 so:
\[\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r} \iff |r|<1\].
Since r=(-1/2)=>|-1/2|<1. The series converges to:
\[\frac{\frac{-1}{2}}{1-(\frac{-1}{2})}\]. or -1/3

Answer 23

Sorry that took me so long to type. Latex can be a feather sometimes xP

Answer 24

Make sense? :O

Answer 25

one sec

Answer 26

OMG! got it now!
Thanks a lot!

Answer 27

No problem :) Just post any questions you have Calculus is my thing xP

Answer 28

maleviolence19 dat was a stud proof !

Answer 29

Haha, Thank you! I try :D

Answer 30

Is sigma n=1 to infinity of (3^n -1) divided by 2^n divergent?
just tell me if it is right or wrong!

Answer 31

yeap!

Answer 32

It is divergent. Because you can break it up:
\[\sum_{n=1}^{\infty}\frac{3^n-1}{2^n}=\frac{3}{2}\sum_{n=1}^{\infty}(\frac{3}{2})^{n-1}-\frac{1}{2}\sum_{n=1}^{\infty}(\frac{1}{2})^{n-1}\].
You know that the left one diverges and the right one converges. But a divergent added to a convergent still diverges!!!

Answer 33

ty!

Answer 34

No problem!

Answer 35

You could also do the ratio test :P

Answer 36

You are so smart! are you a calc prof or sth?

Answer 37

Nah, haha. I am a math major senior in college though :P

Answer 38

Nice!

Answer 39

Yeah :D

Answer 40

Sigma n=1 to infinity (8^n)/(8^n)..diverges right