# Equivalent expressions for infinite series

• MathewsMD
In summary, the conversation is about finding the series representation of ln(x) and whether there are any errors in the proposed methods. The first method involves using the formula ##f(x) = ∑^∞_{n=0} \frac {f^{(n)}(0)}{n!}x^n## to represent ## \frac {1}{x}## as a series, and then integrating to find the series representation for ln(x). The second method involves using the geometric series formula ##\frac {1}{1-r} = ∑^∞_{n=0} r^n## and letting r = x + 1, then integrating to find the series representation for ln(x). The conversation also includes discussions
MathewsMD
Question:

I was just wondering if there was any error in what I've done in the following steps to find the series representation of ##lnx##. I know ## \frac {1}{x}## is given in the following link by doing having the a function centred at 0, you can let ##f(x) = ∑^∞_{n=0} \frac {f^{(n)}(0)}{n!}x^n## after proving the function ## \frac {1}{x} ## can be represented as a series by doing the remainder test and finding R(x) → 0 as n → 0. Then just integrating.

I was just looking at an alternate method, which I don't find used as commonly but seems a lot simpler. Here are the steps:

Relevant Forumla(s):

##\frac {1}{1-r} = ∑^∞_{n=0} r^n## for a = 1

Attempted Solution:

If we let r = x + 1, then:

##\frac {1}{1- (x+1)} = \frac {1}{x} = ∑^∞_{n=0} (x+1)^n##

Then integrate this function with respect to x to find the representation for lnx

## ∑^∞_{n=0} \frac {(x+1)^{n+1}}{n+1} = lnx ##

For R = 1 on x:(-2,0)

Is my radius and interval of convergence correct? Just wondering, what exactly does the interval and radius of convergence represent for lnx in this case if this is the domain for the series? Since lnx is not defined on (-2,0) I am not exactly understanding what this means for f(x).

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Could you use the template and proofread what you post ? What relevant equations do you have available ? Do you have a series representation of ln(1+x) at hand ?

First of all, integrating ##\frac{1}{x}## doesn't give ##\mathrm{ln}(x)##, but ##\mathrm{ln} |x|##.

micromass said:
First of all, integrating ##\frac{1}{x}## doesn't give ##\mathrm{ln}(x)##, but ##\mathrm{ln} |x|##.

Okay. How exactly would the domain of the series be restricted as to only show for (0,∞) for lnx? Is it possible? Just to confirm, would my solution be correct if I was looking at lnlxl = f(x)? Wouldn't the answers for letting x = -1 give me ln1 for example? So if i just let y = -x and sub y in for x, then i would get the interval for (0,2) right? Please correct me where I am going and explain please since I'd like to get all these misconceptions out of my head before they affect my future understanding of this content.

If I'm not mistaken, the interval of convergence for the power series representation is the same interval of f(x) that can be assessed using the power series representation, right?

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MathewsMD said:
I was just wondering if there was any error in what I've done in the following steps to find the series representation of ##lnx##. I know ## \frac {1}{x}## is given in the following link by doing having the a function centred at 0, you can let ##f(x) = ∑^∞_{n=0} \frac {f^{(n)}(0)}{n!}x^n## after proving the function ## \frac {1}{x} ## can be represented as a series by doing the remainder test and finding R(x) → 0 as n → 0. Then just integrating.

I was just looking at an alternate method, which I don't find used as commonly but seems a lot simpler. Here are the steps:

##\frac {1}{1-r} = ∑^∞_{n=0} r^n## for a = 1

If we let r = x + 1, then:

##\frac {1}{1- (x+1)} = \frac {1}{x} = ∑^∞_{n=0} (x+1)^n##

Then integrate this function with respect to x to find the representation for lnx

## ∑^∞_{n=0} \frac {(x+1)^{n+1}}{n+1} = lnx ##

For R = 1 on x:(-2,0)

Is my radius and interval of convergence correct? Just wondering, what exactly does the interval and radius of convergence represent for lnx in this case if this is the domain for the series? Since lnx is not defined on (-2,0) I am not exactly understanding what this means for f(x).

You can make that work, but you are being way too sloppy. ##\frac {1}{1- (x+1)}## isn't equal to ##\frac {1}{x}##, it's ##\frac {1}{-x}##. And the integral of that is -log(|x|)+C. It's probably worth checking that the integration constant is zero. Since x is in (-2,0) to make the geometric series converge, you can write that as -log(-x). That's the function you are deriving a series for.

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MathewsMD said:
Okay. How exactly would the domain of the series be restricted as to only show for (0,∞) for lnx?

You can't find a power series representation for ##\mathrm{ln}(x)## that works on entire ##(0,+\infty)##. However, for any ##L>0##, you can find one that works on ##(0,L)##.

Just to confirm, would my solution be correct if I was looking at lnlxl = f(x)?

No, since you made some arithmetic errors. See Dick's reply. You should obtain

$$\sum_{n=0}^{+\infty} \frac{(-1)^{n+1}}{n+1} = -\mathrm{ln}|x|$$

for ##x\in (-2,0)##.

micromass said:
You can't find a power series representation for ##\mathrm{ln}(x)## that works on entire ##(0,+\infty)##. However, for any ##L>0##, you can find one that works on ##(0,L)##.

No, since you made some arithmetic errors. See Dick's reply. You should obtain

$$\sum_{n=0}^{+\infty} \frac{(-1)^{n+1}}{n+1} = -\mathrm{ln}|x|$$

for ##x\in (-2,0)##.

No, that's not what you should get. There's no x on the left side.

Dick said:
No, that's not what you should get. There's no x on the left side.

Obviously. I'm an idiot.

micromass said:
Obviously. I'm an idiot.

Yes, but an idiot of extraordinary talent! MatthewsMD should try and do better.

Dick said:
You can make that work, but you are being way too sloppy. ##\frac {1}{1- (x+1)}## isn't equal to ##\frac {1}{x}##, it's ##\frac {1}{-x}##. And the integral of that is -log(|x|)+C. It's probably worth checking that the integration constant is zero. Since x is in (-2,0) to make the geometric series converge, you can write that as -log(-x). That's the function you are deriving a series for.

Ahhh okay. So I would have to add a constant of integration and my expression is for ln(-x) on (-2,0).

MathewsMD said:
Ahhh okay. So I would have to add a constant of integration and my expression is for ln(-x) on (-2,0).

The constant of integration turns out to be 0. And your expression isn't for ln(-x). Can you try to be careful for a change and show what you mean?

## 1. What is an infinite series?

An infinite series is a mathematical expression that represents the sum of an infinite number of terms. It is denoted by the symbol Σ (sigma) and has the general form of a1 + a2 + a3 + ... + an, where n is the number of terms and an is the nth term in the series.

## 2. How do you find the sum of an infinite series?

In order to find the sum of an infinite series, you need to determine the limit of the partial sums as the number of terms approaches infinity. This is known as the limit of the series and can be found using various mathematical techniques such as the ratio test, comparison test, or integral test.

## 3. What are equivalent expressions for infinite series?

Equivalent expressions for infinite series are different ways to represent the same sum. For example, the infinite series 1 + 1/2 + 1/4 + 1/8 + ... can also be written as the infinite series 20 + 2-1 + 2-2 + 2-3 + ... Both of these expressions represent the sum of all powers of 2, which is equivalent to 2.

## 4. How do you determine if two infinite series are equivalent?

In order to determine if two infinite series are equivalent, you can use the comparison test. This test compares the terms of the two series and determines if one series is always greater than or equal to the other series. If the terms are always equal, then the two series are equivalent.

## 5. Why are equivalent expressions for infinite series important?

Equivalent expressions for infinite series are important because they allow us to manipulate and simplify complicated sums. By finding equivalent expressions, we can often find an easier or more efficient way to calculate the sum of an infinite series. Additionally, they help us to better understand the properties and behavior of infinite series in mathematics.

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