Analyzing the Convergence of a Geometric Series

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In summary, the conversation discusses finding the sum of a series using different methods, such as expressing the result as a logarithm and using pattern recognition. The possibility of finding the constant of integration and its relation to the initial conditions is also explored.
  • #1
Mr Davis 97
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I have the following series that I came up with in doing a problem: ##\displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}##. I looked at WolframAlpha and it says that this series converges to ##\log (2)##. Is it possible to figure this out analytically?
 
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  • #2
Mr Davis 97 said:
I have the following series that I came up with in doing a problem: ##\displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}##. I looked at WolframAlpha and it says that this series converges to ##\log (2)##. Is it possible to figure this out analytically?
As the result involves the logarithm, we first have to find a way to express the logarithm by a series. Which one do you chose? If you define the logarithm otherwise, there will probably a bit more work to do.
 
  • #3
fresh_42 said:
As the result involves the logarithm, we first have to find a way to express the logarithm by a series. Which one do you chose? If you define the logarithm otherwise, there will probably a bit more work to do.
Well, suppose that I had no idea that the answer turned out to be ##\log 2##. Would there be any way for me to proceed?
 
  • #4
Pattern recognition and a slightly more general problem:

use ## p \in (0,1)## instead of ##\frac{1}{2}##

what happens if you differentiate the series?
 
  • #5
Mr Davis 97 said:
Well, suppose that I had no idea that the answer turned out to be ##\log 2##. Would there be any way for me to proceed?
The easiest way is to consider ##\log(1+x)=-\sum_{k=0}^{\infty}\dfrac{(-x)^{k+1}}{k+1}## which is the power series expansion at ##x=0## for ##-1< x \le 1\,.## Then simply take ##x=-\frac{1}{2}\,.## But I think it's fun (and a good exercise) to consider the other proposed ways above, esp. what @StoneTemplePython suggested: differentiate ##f(p)=\sum_{k=0}^{\infty}\dfrac{p^{k+1}}{k+1}##.
 
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  • #6
StoneTemplePython said:
Pattern recognition and a slightly more general problem:

use ## p \in (0,1)## instead of ##\frac{1}{2}##

what happens if you differentiate the series?
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.
 
  • #7
Mr Davis 97 said:
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.

actually you do know the initial conditions are

##
\displaystyle \sum_{n=0}^{\infty} \frac{p^{n+1}}{(n+1)}
##

so when you have

##\frac{1}{1-p} = 1 + p + p^2 + p^3 + ... ##

you then integrate both sides... ignoring the constant of integration for a moment, does the term by term integration of the Right hand side match with your original series? If so, the Left hand side does too -- this allows you to "zero in" on the constant of integration.

- - - -
edit: another simpler thing to notice is that every term in your original series has a ##p## there are no constants...

another edit: changed the original series to make sure the ##p## is in the numerator!
 
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  • #8
Mr Davis 97 said:
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.
What is ##\sum q^k## for ##0<q<1\,##?
 
  • #9
StoneTemplePython said:
actually you do know the initial conditions are

##
\displaystyle \sum_{n=0}^{\infty} \frac{1}{p^{n+1}(n+1)}
##

so when you have

##\frac{1}{1-p} = 1 + p + p^2 + p^3 + ... ##

you then integrate both sides... ignoring the constant of integration for a moment, does the term by term integration of the Right hand side match with your original series? If so, the Left hand side does too -- this allows you to "zero in" on the constant of integration.

- - - -
edit: another simpler thing to notice is that every term in your original series has a ##p## there are no constants...
So are you saying that the constant of integration is 0?
 
  • #10
Mr Davis 97 said:
So are you saying that the constant of integration is 0?

You tell me. When you applied the derivative operator to your original series, did any term get mapped to zero? That is the information loss associated with basic differentiation. So what kind of information was lost here?
 
  • #11
StoneTemplePython said:
You tell me. When you applied the derivative operator to your original series, did any term get mapped to zero? That is the information loss associated with basic differentiation. So what kind of information was lost here?
I would suppose none... But if ##S(p) = \log p## then ##S(1/2) = -\log 2## which has a sign error
 
  • #13
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Related to Analyzing the Convergence of a Geometric Series

1. What is the definition of a series?

A series is a sequence of numbers that are added together in a specific order.

2. What is the formula for finding the sum of a series?

The formula for finding the sum of a series is S = a + ar + ar^2 + ar^3 + ... + ar^n-1, where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.

3. What is the difference between an arithmetic series and a geometric series?

An arithmetic series is a series where each term is obtained by adding a constant value to the previous term, while a geometric series is a series where each term is obtained by multiplying the previous term by a constant value.

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

The sum of an infinite series can be found by using the formula S = a/(1-r), where S is the sum, a is the first term, and r is the common ratio. This formula only works for geometric series where the absolute value of the common ratio is less than 1.

5. Can the sum of a series be negative?

Yes, the sum of a series can be negative if the terms in the series alternate between positive and negative values. However, if the terms in the series are all positive, then the sum will also be positive.

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