Bit confused about the geometric series

[Bazzinga]

I'm confused about the sum of the geometric series:

\sum ar^{n-1} = \frac{a}{1-r} when |r|<1

but if you have a series like:

\sum (1/4)^{n-1}

the sum is:

\frac{1/4}{1-(1/4)}

should't it be \frac{1}{1-(1/4)} because there is no a value?

 

[tiny-tim]

Hi Bazzinga!

Depends where you start from: n = 1 or n = 2.

 

[Bazzinga]

Well, for the question's sake let's say we start at n=1, why is there a difference if we start at n=2?

 

[tiny-tim]

Well, for the question's sake let's say we start at n=1, why is there a difference if we start at n=2?

erm …

because every term in the second series is 1/4 times the same term in the first series!

 

[Bazzinga]

Not following :S

 

[tiny-tim]

1/4 + 1/16 + 1/64 + …

1/16 + 1/64 + 1/256 + …

the second line is 1/4 times the first line

 

[Bazzinga]

Ok.. so why does my textbook say the sum is a/(1-r) when it's actually ar/(1-r) ?

 

[tiny-tim]

dunno

 

[Bazzinga]

So the textbook is wrong? What is the actual sum of the geometric series then?

 

[Dick]

So the textbook is wrong? What is the actual sum of the geometric series then?

No one can say until you tell us what the limits on the summation are. The textbook really doesn't show any??

 

[gb7nash]

We can't answer anything without knowing what the starting point and ending point is on the sum.

 

[Bazzinga]

how about n=1->infinity? The textbook doesn't say, it just says that's the sum =\ I'm assuming its n=1->inf though

 

[gb7nash]

Well, assuming it starts at one and goes to infinity, then yes, your answer should be \frac{1}{1-(1/4)} = 4/3

You might want to look where your book defines the sigma notation for the sum... It might say somewhere "if no starting and ending point are on the sum, assume _____". If not, that's pretty poor form for a question.

 

[Bazzinga]

Ok I just looked at another question and it does have n=1 to infinity, but it still defines the sum like:
(1/4)/(1-(1/4))

 

[Dick]

Ok I just looked at another question and it does have n=1 to infinity, but it still defines the sum like:
(1/4)/(1-(1/4))

Are you saying it DOESN'T give limits on this problem?? The sum from n=1 to infinity of (1/4)^(n-1) is 1/(1-(1/4)). If that's the problem the book gives then that is the answer. If the books says otherwise, it's wrong. Here's a much better question. Do you know how you would figure out for yourself if the book was wrong? Presuming you can figure out what the question is.

 

[Bazzinga]

No idea how to figure it out, want to help me through it instead of being a dick, Dick?

 

[SammyS]

No idea how to figure it out, want to help me through it instead of being a dick, Dick?

You might want to edit your inappropriate response, that is, if you want to use this website to get any more help.

Note: Dick has 16,513 posts on this site, and he is an official "Homework Helper" and an official "Science Advisor".

 

[Dick]

No idea how to figure it out, want to help me through it instead of being a dick, Dick?

Oh, I think you've probably had enough help. Bye.

 

[Bazzinga]

Keep your panties on ladies, I was making a pun

 

[Dick]

Keep your panties on ladies, I was making a pun

Exactly what kind of help do you need? I can't find the limits for you. They HAVE to be stated in the problem. Why don't you state a problem with explicit limits? Then say why you are having trouble with it.

 

[Bazzinga]

I was hoping someone could help me understand how to find the actual sum of the geometric series, whether it starts at n=0, n=1 or n=N or whatever!

 

[Bazzinga]

But judging by what everyones been saying, I'm assuming if n=1, then the sum is a/(1-r) and if it starts at 0 you have to multiply everything by r, making the actual sum ar/(1-r)

EDIT: n=1 to infinity of course

 

[Dick]

I was hoping someone could help me understand how to find the actual sum of the geometric series, whether it starts at n=0, n=1 or n=N or whatever!

Do you know how to find the sum of a+a*r+a*r^2+a*r^3+...? Let's forget about the summation notation for a minute.

 

[Bazzinga]

So far in class to find the sum we've just been comparing everything to the geometric series or using a bound on the error, so no

 

[Dick]

So far in class to find the sum we've just been comparing everything to the geometric series or using a bound on the error, so no

Ok, then. Here's the classic way to figure it out without a formula. Let S=a+a*r+a*r^2+a*r^3+... What's S-r*S? Cancel a bunch of stuff.

 

[Bazzinga]

Oh! Everything would cancel out except a - ar^n

S - r*S = a - ar^n
...
S = a(1 - r^n)/(1-r)

which explains why -1<r<1 since the limit n->inf r^n would diverge above 1

 

[Dick]

Oh! Everything would cancel out except a - ar^n

S - r*S = a - ar^n
...
S = a(1 - r^n)/(1-r)

which explains why -1<r<1 since the limit n->inf r^n would diverge above 1

Sure. But my '...' was supposed to mean 'to infinity'. That means r^n->0. So you just get a/(1-r), right?

 

[Bazzinga]

Ok, but there's this example in my textbook:

\sum^{n=1}_{infinity}(1/2)^{n}=\frac{1/2}{1-(1/2)}=1

"The series is a geometric series with a=1/2 and r=1/2"

I'm confused as to how a=1/2

 

[Bazzinga]

Oh I think I just got it...

\sum^{n=1}_{infinity}(1/2)(1/2)^{n-1}=\frac{1/2}{1-(1/2)}=1

durr

 

[Dick]

Ok, but there's this example in my textbook:

\sum^{n=1}_{infinity}(1/2)^{n}=\frac{1/2}{1-(1/2)}=1

"The series is a geometric series with a=1/2 and r=1/2"

I'm confused as to how a=1/2

Start writing the sum out. n=1 is 1/2, n=2 is (1/2)^2, n=3 is (1/2)^3. So you've got (1/2)+(1/2)^2+(1/2)^3+... If you put a=(1/2) and r=(1/2) in a+a*r+a*r^2+... you get the same series.

 

[tiny-tim]

(just got up :zzz: …)

Bazzinga, you need to learn these two formulas …

for all n:

1 - xⁿ = (1 - x)(1 + x + x² + … + x^n-1)​

for odd n:

1 + xⁿ = (1 + x)(1 - x + x² - … + x^n-1) ​