Find the sum of this series when n goes from 1 to infinite

[monsmatglad]

((2x-1)^n)/n find the sum of this series when n goes from 1 to infinite
please help

I have tried this several times, but it doesn't match with the true answer.

thank you

 

[Char. Limit]

Try substituting in u=2x - 1 for now, to get \sum_{n=1}^\infty \frac{u^n}{n}, which is easily recognizable as the integral of u^(n-1). Now, do you know the series for u^(n-1)?

 

[monsmatglad]

what I tried was to derivate to get rid of the n under, which then gave me 2*(2x-1)^n-1. Then I used the sum of a geometric series, getting: 2/(1-(2x-1))= 2/(2-2x)=1/(1-x).
Then i integrated back 1/(1-t) with t going from 0 to x. Then I get -ln(1-t), but this does not fit with what the answer should be.

 

[Char. Limit]

Actually, the solution IS correct, up to a constant. The problem is, by deriving, you lost the constant. So all you need to do now is figure out what that constant term is.

 

[monsmatglad]

that's good news:P but I don't see how to find the constant term. The true answer should be -ln(2-2x) .

 

[Char. Limit]

Try factoring that. You get -ln(2 (1 - x)). Using the rules of logarithms, how can you factor that out?

 

[monsmatglad]

-(ln2+ln(1-x)), but i don't see how to derive this without going through the true answer.

 

[monsmatglad]

I mean, how do I get to the constant without using the correct answer to derive from.

 

[monsmatglad]

That might have been a bit fuzzy due to some sloppiness with names, but basically I need to know how to find that constant term that disappears when deriving.

 

[micromass]

So you've figured out correctly that

\sum_{n=1}^{+\infty}{\frac{(2x-1)^n}{n}}=-ln(1-x)+C

So, if you fill in a particular x, then this equality must hold. You might fill in 0 for x, but then you'll have to evaluate the series

\sum_{n=1}^{+\infty}{\frac{(-1)^n}{n}}

which may be a little difficult. Can you think of an x that will make the serie easy? (for example, an x such that all the terms of the series are 0?)

 

[monsmatglad]

x=2?

 

[monsmatglad]

ehm. x=2 can't be correct, but with x=0, you get ln(1)=0
right?

 

[monsmatglad]

-1/2?
sorry, I'm stupid

 

[monsmatglad]

that was supposed to be 1/2 without the minus

 

[micromass]

Will the series be easy if you fill in x=2? You will get the series

\sum_{n=1}^{+\infty}{\frac{3^n}{n}}

which isn't easy to determine.

In fact, you probably have somewhere in your assignment the contraint that 0<x<1. So you can't fill in 2 since the series will diverge for that value!

 

[micromass]

that was supposed to be 1/2 without the minus

Yes! That is good. What do you get for x=1/2?

 

[monsmatglad]

ah, so any value that makes the serie converge will do?

 

[micromass]

ah, so any value that makes the serie converge will do?

Well, yes and no. Every value that makes the series converge will do in the sense that it gives you a value for C.
However, you will want to pick a value for x that makes the serie easy. If you pick x=1/2, then this is the case since the series will vanish. So this is the ideal value to take! You can pick x=3/4 if you want, but the serie will be much more difficult to evaluate!

 

[monsmatglad]

so, i pick 1/2. How do I get from that to getting the constant -ln2?

 

[micromass]

Well, substitute 1/2 in your equation. What do you get?

 

[Char. Limit]

Remember first that you have f(x) = -ln(1 - x) + C, and you want to find C.

 

[monsmatglad]

-ln(1/2)=ln2? that's right isn't it?

 

[monsmatglad]

i think i have got it. i get 0 on one side av ln2 + C on the other. moving them around gives - ln2. correct?

 

[Char. Limit]

That's correct. So now you have your constant term, and can write the answer to the series.

 

[monsmatglad]

thank you very much for being so patient.=)