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Does this need Abel's theorem? watch

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    Prove \displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = 2



    What theorem do I need for this?
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    No theorem needed (theorems would generally tell you whether or not a series converges, but not what it converges to). There's a trick for these sorts of sums. Consider the expansion  (1-x)^{-1} = 1 + x + x^2 + ... . How can you manipulate this to help you? (it might help you to write out your sum more fully). Hint below:

    Spoiler:
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    You're allowed to differentiate/integrate, multiply through by things...
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    (Original post by Daniel Freedman)
    No theorem needed (theorems would generally tell you whether or not a series converges, but not what it converges to). There's a trick for these sorts of sums. Consider the expansion  (1-x)^{-1} = 1 + x + x^2 + ... . How can you manipulate this to help you? (it might help you to write out your sum more fully). Hint below:

    Spoiler:
    Show
    You're allowed to differentiate/integrate, multiply through by things...
    Okay so I got that to be  (1-x)^{-1} = \displaystyle\sum_{n=0}^{\infty} x^n

    I let x = 2 =>  -1 = 1 + \displaystyle\sum_{n=1}^{\infty} 2^n

    How do I move on from this?
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    Also I tried just putting x = 1/2 into the equation you gave but that didn't yield the right result!
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    Firstly, remember that the series  1 + x + x^2 + ... only converges for |x| < 1, so you cant let x = 2.

    Substituting in an appropriate value should be the final thing you do. First, you need to manipulate  1 + x + x^2 + ... to make it look more like your series. What happens if you differentiate it?
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    (Original post by Daniel Freedman)
    First, you need to manipulate  1 + x + x^2 + ... to make it look more like your series. What happens if you differentiate it?
    You need to justify why you can differentiate term-by-term and get the same result as differentiating the expression for the sum.
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    (Original post by Daniel Freedman)
    Firstly, remember that the series  1 + x + x^2 + ... only converges for |x| < 1, so you cant let x = 2.

    Substituting in an appropriate value should be the final thing you do. First, you need to manipulate  1 + x + x^2 + ... to make it look more like your series. What happens if you differentiate it?
    This gets \displaystyle\sum_{n=1}^{\infty} n x^{n-1}

    How do I get this to become (I assume) \displaystyle\sum_{n=1}^{\infty} nx^{-n}?
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    You don't need to. Set x = 1/2.
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    (Original post by DFranklin)
    You don't need to. Set x = 1/2.
    Almost there

    - (1 - \frac{1}{2})^{-2} = 2 \displaystyle\sum_{n=1}^{\infty} n (\frac{1}{2})^{n}

    =>  - 4 = 2 \displaystyle\sum_{n=1}^{\infty} \frac {n}{2^n}

    there seems to be a sign error but I can't find it!
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    You've differentiated wrongly. (You get one minus from the ^-1 exponent, and another because it's (1-x), not (1+x)).

    Note that this is still NOT a valid proof in an analysis sense - you can't differentiate an infinite series term by term without justification.
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    (Original post by Dagnabbit)
    Prove \displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = 2



    What theorem do I need for this?
    Don't you just use proof by mathematical induction?I.e. Let n=1, if true then we can say is also true for n=k, then you prove it works for n=k+1, thus true for any value.
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    (Original post by ConorMC)
    Don't you just use proof by mathematical induction?I.e. Let n=1, if true then we can say is also true for n=k, then you prove it works for n=k+1, thus true for any value.
    No. You can use induction to prove something works for finite cases, but this does not directly carry through to the infinite case.

    In this case, we don't even have a statement of the result for the finite case. A valid way of proving the desired result is to find the result for the finite case, prove it by induction, and then examine what happens as the number of terms goes to infinity. Unfortunately, the finite result is more complicated than the infinite one.
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    (Original post by ConorMC)
    Don't you just use proof by mathematical induction?I.e. Let n=1, if true then we can say is also true for n=k, then you prove it works for n=k+1, thus true for any value.
    We have no way of knowing it's true for n = 1 though!
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    (Original post by DFranklin)
    You've differentiated wrongly. (You get one minus from the ^-1 exponent, and another because it's (1-x), not (1+x)).

    Note that this is still NOT a valid proof in an analysis sense - you can't differentiate an infinite series term by term without justification.
    Ah, yep, cheers muchly!


    Would it help if we assumed the series converged uniformly for |x| < 1?
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    (Original post by Dagnabbit)
    Would it help if we assumed the series converged uniformly for |x| < 1?
    Well, if you're going to assume stuff, why not just assume the result? (You are actually OK if the differentiated series converges uniformly, but it's not incredibly obvious).
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    (Original post by DFranklin)
    Well, if you're going to assume stuff, why not just assume the result? (You are actually OK if the differentiated series converges uniformly, but it's not incredibly obvious).
    I think this is what Abel's theorem might be about, I'll read up on it a bit. Cheers for help
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    (Original post by Dagnabbit)
    I think this is what Abel's theorem might be about, I'll read up on it a bit.
    No, it isn't. (The point you are interested in isn't on the radius of convergence, so you don't need Abel).
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    (Original post by DFranklin)
    No, it isn't. (The point you are interested in isn't on the radius of convergence, so you don't need Abel).
    Is it something that can be tackled in second year analysis?
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    (Original post by Dagnabbit)
    Is it something that can be tackled in second year analysis?
    First year, even.

    If you have the theorem that a power series can be differentiated term by term within it's radius of convergence, you can use that.

    Otherwise, define

    S_n(x) = \sum_{r=1}^n n x^n

    You can find S_n(x) explicitly by considering S_n(x) - xS_n(x). (You get a GP with a couple of extra terms at beginning/end).
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    (Original post by DFranklin)
    First year, even.

    ...
    I used Weierstrass' M-test which seems to work
 
 
 
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