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Series and Future Stock Prices

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a question here that seems so simple, yet I can't seem to wrap my head around what they're asking.

    It says that one method of pricing a stock is to set the share price equal to the sum of all future dividends for infinitely many years, with dividends discounted to their present value. We assume that we are always able to invest money at a guaranteed rate of 5% per year. This implies that $1.05 received next year is worth only $1 today. What is $1 received n years from now worth today?

    The second part says that if a corporation promises to pay a dividend of $1 per share every year for all years in the future, what is the total value of all future dividends for one share for infinitely many years discounted to their value today.

    2. Relevant equations

    Possibly the compound interest formula?
    A = P( 1 + r )n

    3. The attempt at a solution
    I am assuming the equation for the first part will have something to do with the compound interest formula, but I'm not sure how to apply it here. I was thinking of something like 1 - .05n, but that doesn't seem quite right to me.

    As for the second part of the problem, it is dependent on what I find for the first part, but I think that it has to do with finding the limit of whatever I find to be the answer for the first part of the question.

    Any suggestions that will point me in the right direction here?
     
  2. jcsd
  3. Jan 23, 2013 #2

    HallsofIvy

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    Yes, with "P" the future value and A= $1. Assuming r= 0.05, that gives 1= P(1.05n) so that P= 1/(1.05n).

    For the second part, sum over all n:
    [tex]\sum_{n=0}^\infty \frac{1}{1.05^n}[/tex]
    That is a "geometric series" of the form [itex]\sum r^n[/itex] with r= 1/1.05.
     
  4. Jan 23, 2013 #3
    Hm, I have only seen geometric series of the form [itex]\Sigma[/itex] Arn-1. I understand how to manipulate a term to make a power of n a power of n-1, but the exponent here would be a negative n. Should I instead to be looking at the series as [itex]\sum\left(\frac{1}{1.05}\right)^{n}[/itex] and then manipulate the series to get that power to n-1?
     
  5. Jan 23, 2013 #4

    Ray Vickson

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    What do YOU think?
     
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