- #1

- 1,462

- 44

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

- 16,462

- 15,548

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

- 1,462

- 44

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?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.

- #4

StoneTemplePython

Science Advisor

Gold Member

- 1,203

- 597

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

what happens if you differentiate the series?

- #5

- 16,462

- 15,548

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}##.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?

Last edited:

- #6

- 1,462

- 44

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##.

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

what happens if you differentiate the series?

- #7

StoneTemplePython

Science Advisor

Gold Member

- 1,203

- 597

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.

- - - -

Last edited:

- #8

- 16,462

- 15,548

What is ##\sum q^k## for ##0<q<1\,##?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##.

- #9

- 1,462

- 44

So are you saying that the constant of integration is 0?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 areno constants...

- #10

StoneTemplePython

Science Advisor

Gold Member

- 1,203

- 597

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

- 1,462

- 44

I would suppose none... But if ##S(p) = \log p## then ##S(1/2) = -\log 2## which has a sign errorYou 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?

- #12

StoneTemplePython

Science Advisor

Gold Member

- 1,203

- 597

http://www.wolframalpha.com/input/?i=integral+1/(1-p)I would suppose none... But if ##S(p) = \log p## then ##S(1/2) = -\log 2## which has a sign error

- #13

- 1,462

- 44

Ah! I was doing basic calculus wrong, my bad. I completely get it now, thanks!

Share: