Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Infinite Sum of Fractions
  1. Infinite sum (Replies: 31)

  2. Fractional part sum (Replies: 7)