Question

Find the sum of the series and what values of the variable does the series converge?

y-y^2+y^3-y^4....

A1=y
r=-y
n=4

Can i use A1=(1-r^n/1-r) as the formula? and how do i tell if it converges?

Answer 1

note that here, there is "..."
what does that mean ?

Answer 2

im not sure thats just whats written in my book

Answer 3

more terms?

Answer 4

if you look at your last question, .....anyways
this means its an infinite series, there are no finite number of terms

Answer 5

and sum of infinite geometric series is
\(\Huge S_\infty=\dfrac{a}{1-r}\)
a= 1st term

Answer 6

so, what sum do u get here ?

Answer 7

so do i need to do that with the previous problem too?

Answer 8

y/1+y

Answer 9

no, not with previos problem, it had finite terms, =6
and yes, that sum is correct.
now, any idea on how to find whether it converges ?

Answer 10

take the limit as it approaches infinity?

Answer 11

yes, thats one of the test, you can use it,
there are other tests too, are you taught any of those ?

else you can directly think logically that if y >1 , the sum will go on increasing as number of terms increases, and for n->infinity, sum will also be infinite.
hence that sequence converges only if y <1
makes sense ?

Answer 12

i get that y>1 when n approaches infinity but i dont get how it converges only if y<1

Answer 13

sum converges means sum is finite when n->infinity
sum diverges means sum is infinite when n->infinity

Answer 14

and here sum will be finite only if y<1

Answer 15

have you taught the ratio test to find whether the sequence is convergent ?

Answer 16

http://en.wikipedia.org/wiki/Ratio_test
see "The TEST " part.

Answer 17

if you planing to use it, you need to get the general term a_n

Answer 18

ive heard the ratio test in class but my professor doesnt speak english very well and all he does it write random things on the board and i have no idea whats going on half the time

Answer 19

i think one of the random thing he wrote was the formula
\(\Large \lim \limits_{n \rightarrow \infty } |\dfrac{a_{n+1}}{a_n}|\)

Answer 20

if he had really taught you the ratio test

Answer 21

lets go for it,
do u know how to find the general term of your sequence ?

Answer 22

ive seen something like that!

Answer 23

yes:)

Answer 24

first term in the sequence

Answer 25

general term means you represent the sequence in terms of 'n' ,
like
1,x,x^2,x^3,x^4 ...
will have general term as
x^{n-1}

because 1st term, ----> n=1 ----> x^0 = 1
2nd term, ---->n=2 ---->x^1 =x
and so on,
got this ?

Answer 26

yup:)

Answer 27

can u try to find general term for your seq ?

Answer 28

y^(n+1/2)? :/ maybe

Answer 29

to verify just put n=1 and see whther you get the first term or not

Answer 30

then n=2 to see whether you get 2nd term or not

Answer 31

im sorry not (1/2) it would be y^(n+y)?

Answer 32

okay i think im just confusing myself

Answer 33

lets forget about the alternate = and - signs for now
y,y^2,y^3,y^4...
would y^n be correct ?

Answer 34

alternate *** + and -

Answer 35

i dont think so

Answer 36

lets check,
y^n, n=1, y
y^n, n=2, y^2
y^n,n=3, y^3
.....
why not ?

Answer 37

jk its right

Answer 38

cool, now lets take the + and - part,
whats (-1)^(n+1) for n=1 ?

Answer 39

and for n=2 ? n=3 ? and so on ?

Answer 40

1

Answer 41

and for n=2 ? n=3 ? and so on ?

Answer 42

-1

Answer 43

1

Answer 44

good! do you realize we get alternate +1 and -1 with that ?

Answer 45

yes:)

Answer 46

so whenever you see alternate + and - , think of using (-1)^(n+1)

so, the general term of our sequence is \(\Large (-1)^{n+1}y^n\)
agree ?

Answer 47

yes

Answer 48

good, thats a_n
can you now find a_(n+1)
?

Answer 49

to find a_n+1 just replace every 'n' in a_n by 'n+1'

Answer 50

already got bored with this?! :P
don't worry, we are more than half way through...stay with me megan!