Question

∑n=1∞(cosnΠ)/2n..converges or diverges?please explain!

Answer 1

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

Answer 2

cos(nPi)=(-1)^(n)

so you have an alternating series that is monotone decreasing, so it should converge

Answer 3

diverges

Answer 4

take rsvitale answer!

Answer 5

so what happens to 2^n ??

Answer 6

is it /2n or /2^n ?

Answer 7

we have an alternating series, so 1/2^n is the "other part" it is always positive and monotone decreasing.

Answer 8

/(2n) and /(2^n) both should converge

Answer 9

what would be the sum and please explain in steps!

Answer 10

ok do you know about geometric series?

Answer 11

yes..is 1/2^n geometric?

Answer 12

we split the numerator and denominator into two sums right?

Answer 13

yes, but some of the terms are negative and some are positive so we need two sums

Answer 14

and then what would be the sum to infinity for cos n pi??

Answer 15

no you dont split numerator and denominator

Answer 16

the terms will be -1/2+1/4-1/8+1/16. we write the negative terms (n is odd) as one sum and the positive terms (n is even) as another sum and add them

Answer 17

ok..and then?

Answer 18

what would be the sum for entire series?

Answer 19

the negative terms:

\[-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{4^n}\]

Answer 20

@rsvitale: Hey! thanks ..understood!

Answer 21

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 22

positive terms:

\[\sum_{n=1}^{\infty}\frac{1}{4^n}\]

you should get -1/3 as the total sum when you add the two together

Answer 23

let me look for a sec

Answer 24

this?:

\[\sum_{n=1}^{\infty}\frac{3^n-1}{2^n}\]

Answer 25

yes!

Answer 26

that is the same as:

\[\sum_{n=1}^{\infty}\left( \frac{3}{2} \right) ^n-\sum_{n=1}^{\infty}\left( \frac{1}{2} \right)^n\]

Answer 27

right?

Answer 28

the first diverges since it is a geometric series with term greater than or equal to 1, So the whole thing diverges