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Closed Form from Summation

  1. Sep 13, 2010 #1

    Ignore the above, I was haveing problems with the symbol...

    Convert each to closed form:

    1. Sum from i=1 to n of: [tex]\frac{n}{a^n}[/tex]

    2. Sum from i=1 to n of: [tex]\frac{1}{a^n}[/tex]


    P.S. I know how to do it if it was an infinite series, but not for this.
  2. jcsd
  3. Sep 13, 2010 #2
    You almost certainly do not mean to ask for the sum of 1/a^n+1/a^n+...+1/a^n adding n copies of 1/a^n.

    Do you mean to ask for the sum of 1/1^n+1/2^n+...+1/n^n?

    Do you mean to ask for the sum of 1/a^1+1/a^2+...+1/a^n?
  4. Sep 13, 2010 #3
    Hi Bill, you are right.

    For #1, I am asking for a closed form equation for the sum of: 1/a^1 + 1/a^2 + 1/a^3 ... 1/a^n

    For #2, I was asking for a closed form equation of the sum of: 1/a^1 + 2/a^2 + 3/a^3 ... n/a^n

    For both, a is any constant (for my purposes a is (1+r) where r is going to be the yield to maturity as I am trying to come up with a closed form equation for the duration of a bond), but I don't think that is relevent to the above.

  5. Sep 13, 2010 #4


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    The first is almost a geometric series- [tex]\sum_{i= 0}^n b^i[/tex] which has a well known sum: [tex]\frac{1}{1- b}[/tex] for -1< b< 1. I say "almost" because it is missing the first term, [itex]b^0= 1[/itex] and has b= 1/a. Of course, that means your sum is just the geometric series without that first "1". Its sum is [tex]\frac{1}{1- b}- 1= \frac{b}{1- b}[/tex], again, for -1< b< 1. And, with b= 1/a, that is [tex]\frac{\frac{1}{a}}{1- \frac{1}{a}}= \frac{1}{a- 1}[/tex] after you multiply both numerator and denominator by a.

    For the second one, think of the series [tex]\sum_{n=1}^\infty \left(\frac{x}{a}\right)^n[/tex] which is just the same as above except with [tex]\frac{x}{a}[/itex] instead of b. It sum is [tex]\frac{\frac{x}{a}}{1- \frac{x}{a}}= \frac{ax}{a- x}[/tex], again multiplying numerator and denominator by a.

    Now look what happens when you differentiate that sum "term by "term". The derivative of [tex]\left(\frac{x}{a}\right)^n[/tex] is [tex]\frac{nx^{n-1}}{a^n}[/tex] and the derivative of [tex]\frac{ax}{a- x}[/tex] is [tex]\frac{a}{(a- x)^2}[/tex].

    That is, [tex]\sum_{n=1}^\infty \frac{nx^{n-1}}{a^n}= \frac{a}{(a- x)^2}[/tex]. Now take x= 1 to get your sum.
  6. Sep 13, 2010 #5


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    The finite sum [tex] \sum_{i= 0}^n b^i [/tex] is (1-bn+1)/(1-b).

    To get your answer let b=1/a.
  7. Sep 14, 2010 #6
    Thanks for the help. Mathman, that was what I was looking for for the first one.

    Using your example, do you or anyone else know what the finite sum would be for i bi?
  8. Sep 14, 2010 #7
    I pasted a handy summation table into my math handbook. I attached a copy for you.

    By the way, you mentioned that you know the formulas for the infinite summations, but not the finite summations. You can actually derive the finite sums from the infinite sums easily. I give an example below.

    [tex]\sum_{k=0}^n a^k=\sum_{k=0}^\infty a^k - \sum_{k=n+1}^\infty a^k [/tex]

    let m=k-n-1 in the second infinite summation.

    [tex]=\sum_{k=0}^\infty a^k - \sum_{m=0}^\infty a^{m+n+1} = \sum_{k=0}^\infty a^k - a^{n+1}\sum_{m=0}^\infty a^m={{1-a^{n+1}}\over{1-a}}[/tex]

    Attached Files:

    Last edited: Sep 14, 2010
  9. Sep 14, 2010 #8


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    Note to stevenb: The infinite sum rule is derived from the finite sum rule.

    Finite sum rule derivation:
    Infinite sum rule is obtained by n->∞ for |a|<1.
  10. Sep 14, 2010 #9
    OK, I don't disagree, but maybe you missed my point.

    The OP said that he knew the infinite sum formulas, but didn't know the finite sum formulas. I was just pointing out to him that he can derive the finite sum formulas from the infinite sum formulas. Are you saying this isn't true, or are you just pointing out an accepted convention?

    I can easily prove the infinite sum formula by doing long division on 1/(1-a), and above I show how to get the finite sum formula from the infinite sum formula. Math is pretty flexible, as I'm sure you know.
  11. Sep 15, 2010 #10


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    The finite sum rule may be derived from the infinite sum rule as you said. The point I was trying to make is that the infinite sum rule is itself derived from the finite rule, which is derived as I expressed.
  12. Sep 15, 2010 #11
    I guess I (politely) object to your phrasing here a little. - no offense intended.

    I find this statement unnecessarily rigid.

    Surely, you can (or may) derive the infinite sum rule from the finite sum rule. Perhaps, it is even preferred (simplest, most elegant?) to do it that way. I don't disagree with that at all. Maybe it was even historically found that way? I don't know the history, so I can't answer that. I would in fact be interested in the history if this is your point.

    However, I'm not aware of any requirement to do it that way. I should be able to choose to derive the infinite sum rule by a number of different methods, some more complicated or obscure than others. I might do that without ever mentioning or thinking about the finite sum formula.

    Just to give some examples with minimal thought.

    1. Do long division on 1/(1-a) and it equals 1+a+a^2 ...

    2. A round about way is to prove the Binomial series formula and then apply it to (1-a)^(-1)

    or mimicking your method for finite sums

    3. Multiply 1+a+a^2 ... times (1-a) and it's clear this equals unity. One can just look at the series and come up with that solution instantly. This is essentially identical to your proof for the finite series case, but using infinite series, where one never needs to think about a finite sum.

    Infinite sum rule derivation:
    S=Σ(k=0,inf) a^k
    aS=Σ(k=1,inf) a^k

    To me, it's just a matter of preference on how you want to think about it. Is there some subtle mathematical reason why one must do it for finite series and then take the limit? Maybe it's one of those finer points that just goes over the head of an engineer?
    Last edited: Sep 15, 2010
  13. Sep 16, 2010 #12


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    The only subtlety comes from the need to include the limit |a|<1 in the discussion. By doing the finite sum first, the answer to this question is obvious, since the discarded term -> 0 only for |a| < 1.
  14. Sep 16, 2010 #13
    Yes, that's a good point. Thank you!
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