Question

Calculus II question regarding series, posted below in a minute.

Answer 1

For the nth partial sum of a geometric series\[\frac{ a }{ (1-r) }\]
where a and r are two constants, egh, let me just attach the specific example, one sec.

Answer 2

(Sorry guys, trying to put this into words, may be a little)

Answer 3

Take your time.

Answer 4

(Sorry guys, trying to put this into words, may be a little)

Answer 5

Webpage just crashed, have to rewrite it -_-

Answer 6

When you have a general term for a sequence that's geometric, and you're looking to find whether the series converges or diverges (general term for a sequence like this:) \[\frac{ (-1)^{n} }{ 4^{n} }\]
and the nth partial sum for a geometric series is \[ar ^{n-1}\]

(cont'd)

Answer 7

Is it safe to assume that the r term is the coefficient for any combined terms with n as an exponent? It seems like every time I see a problem and you have two separate integers or something in the numerator or denominator, they just get consolidated to be (-1/4)^n and r is assumed to be -1/4.

Answer 8

\[\frac{ (-1)^{n} }{ 4^{n} }=(\frac{ -1 }{ 4 })^{n}\]And yes, r is -1/4.

Answer 9

Like if I had \[\frac{ 5(-1)^{n} }{ 3^{n} }\]
r would consequently b (-1/3)?

Answer 10

*be

Answer 11

It's called the \(base\) of the exponent, not the coefficient.

Answer 12

Its just that you can factor out the exponent. Sometimes you can do a little bit of manipulation to make it look like a geo-series. Thats all this is. The term that is being taken to the nth power is r. And yeah, you could do that, say r is -1/3

Answer 13

wio: Sorry, I'm just butchering my math language; I know the proper terms but I'm prone to just getting my point across, my bad.

Answer 14

It's just an exponent rule: \[
a^c\times b^c=(a\times b)^c
\]If you let \(d=\frac 1b\)\[
\frac{a^c}{d^c}=a^c\times \frac 1{d^c}=a^c\times b^c
\]

Answer 15

Yes, I'm quite aware of that, I was just a little unsure about how r was defined and didn't get with certainty what n was with relation to r.

Answer 16

Yeah, basically its the portion you are taking to the nth power. Hopefully a term that can be manipulated such that |r| is between 0 and 1.

Answer 17

Okay, cool. Thanks, guys.

Answer 18

Yep yep.

Answer 19

\(\color{blue}{\text{Originally Posted by}}\) @Mendicant_Bias
For the nth partial sum of a geometric series\[\frac{ a }{ (1-r) }\]\(\color{blue}{\text{End of Quote}}\)
This is not a partial sum. This is the infinite sum's convergence, I believe.
It works when \(|r|<1\) I think.

Answer 20

Otherwise it diverges.

Answer 21

It does work for that; Just learned this yesterday, so my language of stuff like sequences and series is *definitely* not going to be good; I frankly do need a little help clarifying on that, because the terminology more than the math is messing me up.

Answer 22

Then what is the nth partial sum of a geometric series? Is there a generalized partial sum for all geometric series, or no? Because I'm having trouble separating nth partial sums from the sums of sequences and all of the "general" terms that describe a sequence, a series, etc.

Answer 23

A \(sequence\) is basically like a function, but the inputs are only natural numbers. That is the inputs are \(1,2,3,\ldots\) Functions allows any input, though typically in calculus the focus is with all real numbers.

A \(series\) is the sum of a sequence, and it typically refers to the infinite sum.

A \(partial\ sum\) is a sum up to a finite number of terms, so like up to 3 terms or 400 terms, but not infinite terms.

Answer 24

The partial sum of the geometric series comes form multiplying it by \(1-r\), which makes it a telescoping series.

Answer 25

So you start out with \[
\sum_{k=0}^na_1r^{k-1} = a_1\sum_{k=0}^nr^{k-1}
\]

Answer 26

You multiply this by \(1-r\)\[
(1-r)\times a_1\sum_{k=1}^nr^{k-1} = a_1\sum_{k=1}^nr^{k-1} -r^{k}
\]This is a telescoping series \((r^0-\cancel{r^1})+(\cancel{r^1}-r^2)\):\[
a_1(1-r^n)
\]

Answer 27

Now you can divide out the \(1-r\) and get\[
\sum_{k=1}^na_1r^{k-1}=a_1\left(\frac{1-r^n}{1-r}\right)
\]

Answer 28

So, just for my own clarification, the terms to the right of the sigma sign in these instances are the general terms for the sequences that are being summed infinitely.

Answer 29

This partial sum make it easier to test where geometric series converges, because the infinite sum is just the partial sum as \(n\to\infty\)

Answer 30

Yes, that is the correct way of phrasing it. The index is \(n\) so the give the \(n\)th term of the sequence that is being summed.

Answer 31

Okay, awesome, Thank you so much!

Answer 32

Looking back at this, I'm a little confused how you came to the nth partial sum for a geometric series. Why (or, how did you know to?) did you multiply and divide by (1-r) and why did the k's in the exponents of the r terms in the numerator just disappear?

Answer 33

Nevermind, I understand why you multiplied by (1-r) and why the k's disappeared when you showed that it was a telescoping series, what I still don't get (may just take a minute) is why you eventually divided by 1-r. Was that just to compensate for multiplying by it originally?

Answer 34

Also, how does \[\lim_{n \rightarrow \infty}a _{1}\frac{ (1-r ^{n}) }{ (1-r) } = \frac{ a }{ 1-r}\]?

"...the infinite sum is just the partial sum as n→∞", if I took the limit of the above initial term, shouldn't I get infinity?

Answer 35

Well \[
n\to \infty \implies 2^n \to \infty
\]But:\[
n\to \infty \implies \left(\frac{999}{1000}\right)^n\to L\lt \infty
\]

Answer 36

Actually \(L=0\)

Answer 37

I'm a little confused about how you moved n out, this might just be being terrible with exponents, but n was only in the denominator; I thought you could only move the exponent out if both terms (assuming only one in numerator, one in denominator) are to some nth power.

Answer 38

(But yeah, I can see why that limit would equal that.)

Answer 39

*might just be me being terrible with exponents

Answer 40

Well, remember that it doesn't always converge, sometimes it diverges.

Answer 41

There are conditions for convergence.

Answer 42

Yeah, but just purely algebraically, I don't understand how you took the n out.

Answer 43

How do you know \(r\neq 999/1000\)?

Answer 44

Because n was only in the numerator. ?

Answer 45

I'm taking about \(r\).

Answer 46

I don't, I'm not concerned at all with r at the moment, lol, right now I'm just concerned with the exponent, xD

Answer 47

There are cases where \(r^n\to 0\) and where \(r^n\to \infty\).

Answer 48

I know, I'm just talking about the exponent, I just don't get how you're pulling n outside of the single term in the numerator with n involved; nothing else in there is to an nth power, so I don't understand how you're pulling n out in order to take the limit.

Answer 49

Okay, so, are we not taking the limit by moving n outside of the entire thing? I understand exactly what you're saying about if r is less than one and greater than zero, r^n lim n -> infinity will go to zero, but I'm just talking about the algebra of moving n outside.

Answer 50

Nevermind, just a big misunderstanding with what you were saying with the arrows; I haven't seen them used that way, I get what you're saying now.

Answer 51

\[\lim_{n \rightarrow \infty}a _{1}\frac{ \left(1-\left(\frac{999}{100}\right) ^{n}\right) }{ \left(1-\left(\frac{999}{100}\right)\right) } = \frac{ a_1 }{ 1-\left(\frac{999}{100}\right)}\]

Answer 52

Yup.

Answer 53

Sorry about that.