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- Thread starter Mr Davis 97
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fresh_42

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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.

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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.

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StoneTemplePython

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

what happens if you differentiate the series?

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fresh_42

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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?

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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?

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StoneTemplePython

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actually you do know the initial conditions areThat'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##.

##

\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.

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fresh_42

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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##.

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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.

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edit: another simpler thing to notice is that every term in your original series has a ##p## there areno constants...

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StoneTemplePython

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

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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?

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StoneTemplePython

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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

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

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