# Infinite Sum of Fractions

1. Sep 2, 2008

### miggimig

Hi,

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:

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

where 'itex]\sigma[/itex] and $\lambda$ 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:

$\sum_{i=1}^{n}{\frac{1}{a-ib}}$, 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 $\infty$.

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

Thanks a lot,

Michael

2. Sep 2, 2008

### mathman

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

3. Sep 2, 2008

### Hurkyl

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

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)....

4. Sep 6, 2008

### acarchau

$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n(1 + \lambda + \sigma^2)-i(1+\lambda)}\\ =\lim_{n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \frac{1}{(1 + \lambda + \sigma^2)-\frac{i}{n}(1+\lambda)}\\ =\int_{0}^{1}\frac{dx}{(1 + \lambda + \sigma^2)-x(1+\lambda)}$$

5. Sep 7, 2008

### acarchau

The above sum is similar to a difference of harmonic series. Something like
$$H_{2n} - H_{n}=\sum_{k=1}^{n} \frac{1}{n+k}$$
which converges to $$\ln 2.$$

6. Sep 7, 2008

### miggimig

Thank you all very much! You helped me a lot!!!