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Infinite Sum of Fractions

  1. Sep 2, 2008 #1

    actually, I need to calculate an infinite sum of fractions. The problem is that the Limit of the sum is part of the summands. The formula looks like this:

    [tex]\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n(1 + \lambda + \sigma^2)-i(1+\lambda)}[/tex],

    where 'itex]\sigma[/itex] and [itex]\lambda[/itex] are constants. Numerically, this infinite sum converges to a value that can be interpreted as a first order approximation of a channel capacity for some communication scheme.

    My idea to determine this limit is to calculate the finite sum of n terms first. Since, in this case, n is constant, the sum can be written as:

    [itex]\sum_{i=1}^{n}{\frac{1}{a-ib}}[/itex], where a > n b

    When it would be possible to find a conversion of this sum, I thought it might also be possible to determine the limit for n to [itex]\infty[/itex].

    If anyone has ideas how to solve the problem, I would be grateful for comments and feedback.

    Thanks a lot,

  2. jcsd
  3. Sep 2, 2008 #2


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    Are you sure the sum converges? Off hand it looks like (term by term) it converges to a harmonc series, which diverges.
  4. Sep 2, 2008 #3


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    The largest summand is [itex]1 / ( n(1+\lambda + \sigma^2) - (1 + \lambda))[/itex] and the smallest summand is [itex]1 / (n \sigma^2)[/itex], so the limit (or limit points, if it doesn't converge) must be between [itex]1 / \sigma^2[/itex] and [itex]1 / (1 + \lambda + \sigma^2)[/itex], so the limit can't diverge to [itex]\infty[/itex].

    For the opening poster -- it is probably much easier to approximate rather than compute exactly. Since the series does resemble the harmonic series, it might be useful to use a well-known approximation (or one of your basic approximation methods if you don't recall it)....
  5. Sep 6, 2008 #4
    [tex]\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n(1 + \lambda + \sigma^2)-i(1+\lambda)}\\
    =\lim_{n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \frac{1}{(1 + \lambda + \sigma^2)-\frac{i}{n}(1+\lambda)}\\
    =\int_{0}^{1}\frac{dx}{(1 + \lambda + \sigma^2)-x(1+\lambda)}
  6. Sep 7, 2008 #5
    The above sum is similar to a difference of harmonic series. Something like
    [tex]H_{2n} - H_{n}=\sum_{k=1}^{n} \frac{1}{n+k}[/tex]
    which converges to [tex]\ln 2.[/tex]
  7. Sep 7, 2008 #6
    Thank you all very much! You helped me a lot!!!
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