# Can Variable Coefficients Be Used in Geometric Series Sums?

• RJLiberator
In summary: You're right, it's a lot easier to see what is happening if we use x in place of 1/2. So, let's try that:S(x) = -x^3 + 2 x^4 - 3 x^5 + x^6 + x^7 + \cdotsTry writing ##S(x) = -x^3 \, T(x)##, and write out the first few terms of ##T(x)##. Do you now see what is happening?Yes, I do. T(x) is just a geometric series, with the first few terms being -x^2+2x^3-3x^4.
RJLiberator
Gold Member

## Homework Statement

I am giving the sum:
k=1 to infinity Σ(n(-1)^n)/(2^(n+1)

## Homework Equations

first term/(1-r) = sum for a geometric series

## The Attempt at a Solution

[/B]
With some manipulation of the denominator 2^(n+1) = 2*2^n I get the common ratio to be (-1/2)^n while the coefficient is k/2.
The first term is -1/4. This i am confident in.

When I apply the relevant equation, my answer is -1/6.

When I use wolfram alpha calculator the answer is -1/9.

There seems to be something wrong with my manipulation, I have a few guessed:

n/2*(-1/2)^n is my manipulation.

Is it possible to have a variable on the outside of the ratio when applying the geometric series sum? Does the geometric series sum even apply to a problem like this?

Thank you.

RJLiberator said:

## Homework Statement

I am giving the sum:
k=1 to infinity Σ(n(-1)^n)/(2^(n+1)
The index probably shouldn't be k, since the terms being summed all involve n.

In any case, this is not a geometric series.
RJLiberator said:

## Homework Equations

first term/(1-r) = sum for a geometric series

## The Attempt at a Solution

[/B]
With some manipulation of the denominator 2^(n+1) = 2*2^n I get the common ratio to be (-1/2)^n while the coefficient is k/2.
The first term is -1/4. This i am confident in.

When I apply the relevant equation, my answer is -1/6.

When I use wolfram alpha calculator the answer is -1/9.

There seems to be something wrong with my manipulation, I have a few guessed:

n/2*(-1/2)^n is my manipulation.

Is it possible to have a variable on the outside of the ratio when applying the geometric series sum? Does the geometric series sum even apply to a problem like this?

Thank you.

RJLiberator
The index probably shouldn't be k, since the terms being summed all involve n.

In any case, this is not a geometric series.

Damn.

I am guessing that this is NOT a geometric series because it depends on a variable outside the common ratio (k)?

I am also concluding that I need to use the alternating series sum?

RJLiberator said:
Damn.

I am guessing that this is NOT a geometric series because it depends on a variable outside the common ratio (k)?
For a geometric series, an = r*an - 1. You don't have that with the series in this problem.
RJLiberator said:
I am also concluding that I need to use the alternating series sum?
It is an alternating series, yes, and you can get an approximation to the series by adding the first few terms in the series. I think that's what you're talking about.

RJLiberator
Hm. An approximation adding the first few terms...

How would I go about finding the sum (as exact as possible) using calculus II methods?

Nothing
RJLiberator said:
Hm. An approximation adding the first few terms...

How would I go about finding the sum (as exact as possible) using calculus II methods?
Nothing comes to mind. What's the exact wording of the problem? For many of these kinds of problems, they want you to estimate the sum, accurate to, say, four or five decimal places.

RJLiberator said:
Hm. An approximation adding the first few terms...

How would I go about finding the sum (as exact as possible) using calculus II methods?

First, change the problem to ##S(x) = \sum_{k=1}^{\infty} (-1)^k k x^{k+1}##; you want ##S(1/2)##, but it is a lot easier to see what is happening if you use ##x## in place of ##1/2##. So, write out the first few terms of ##S(x)##:
$$S(x) = -x^3 + 2 x^4 - 3 x^5 + \cdots$$
Try writing ##S(x) = -x^3 \, T(x)##, and write out the first few terms of ##T(x)##. Do you now see what is happening?

RJLiberator
@Ray Vickson

Why do we want S(1/2) when the interval is -1<x<1 ?

Wouldn't that s(x) that you wrote out be represented by the first few terms of
-x^2+2x^3-3x^4 and so on and not what you wrote? Or am I missing something here.

Unfortunately, I am not sure what is going on with T(x). I will look more into it when I have time tonight, but if you can offer guidance (or a link) that would be great. :)

Ray Vickson said:
First, change the problem to ##S(x) = \sum_{k=1}^{\infty} (-1)^k k x^{k+1}##; you want ##S(1/2)##, but it is a lot easier to see what is happening if you use ##x## in place of ##1/2##. So, write out the first few terms of ##S(x)##:
$$S(x) = -x^3 + 2 x^4 - 3 x^5 + \cdots$$
Try writing ##S(x) = -x^3 \, T(x)##, and write out the first few terms of ##T(x)##. Do you now see what is happening?

RJLiberator said:
@Ray Vickson

Why do we want S(1/2) when the interval is -1<x<1 ?

Wouldn't that s(x) that you wrote out be represented by the first few terms of
-x^2+2x^3-3x^4 and so on and not what you wrote? Or am I missing something here.

Unfortunately, I am not sure what is going on with T(x). I will look more into it when I have time tonight, but if you can offer guidance (or a link) that would be great. :)

Well, YOU wrote ##\sum_{k=1}^{\infty} (-1)^k k (1/2)^{k+1}##, and that is just ##S(1/2)##.

As to ##T(x)##: the only guidance I am willing to offer is to tell you to sit down and actually write things out. Don't waste your time looking for links or other advice---just do it.

And YES, it should be -x^2 + 2 x^3 - ..., not what I wrote. So S(x) = -x^2 * T(x).

Last edited:
RJLiberator

## 1. What is a geometric series sum?

A geometric series sum is the sum of all terms in a geometric sequence. A geometric sequence is a sequence of numbers where each term is obtained by multiplying the previous term by a constant number, called the common ratio.

## 2. How do you find the sum of a geometric series?

To find the sum of a geometric series, you can use the formula Sn = a1(1 - rn) / (1 - r), where Sn is the sum of the first n terms, a1 is the first term, and r is the common ratio. Alternatively, you can use the formula Sn = a1(rn - 1) / (r - 1).

## 3. What is the common ratio in a geometric series?

The common ratio in a geometric series is the number that is multiplied to each term to get the next term. It is denoted by the letter r and is a constant number.

## 4. Can a geometric series have an infinite sum?

Yes, a geometric series can have an infinite sum if the absolute value of the common ratio is less than 1. In this case, the sum would approach a finite value as the number of terms increases.

## 5. What are some real-life applications of geometric series sums?

Geometric series sums are used in various areas of mathematics, physics, and engineering. Some real-life applications include calculating compound interest, modeling population growth, and analyzing the behavior of electrical circuits.

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