# Is there a closed form of this expression?

1. ### blahdeblah

9
Hi,
(not homework/academic)
Is a closed form of the following expression possible? Either way, some pointers in the right direction would be really helpful.

$$H(s)=\sum_{n=-\infty}^\infty \frac{k^n}{k^n+a/s}$$

Thanks,
D

2. ### Office_Shredder

4,487
Staff Emeritus
Are a, s and k numbers? If so then this sum is divergent - as n goes to +/- infinity, the summand converges to 1 depending on whether k is larger than or smaller than 1

1 person likes this.
3. ### blahdeblah

9
Sorry, the process of posting this, made me think of something which might be helpful:
$$H(s)=\sum_{n=-\infty}^\infty \frac{1}{1+a/(s.k^n)}$$
so...
$$H(s)=\sum_{n=-\infty}^\infty \frac{1}{1+a.k^{-n}/s}$$
I might be able to google this one as it looks a bit more like a standard form of something.

4. ### blahdeblah

9
a is a constant, yes and s is a variable (actually frequency in my application).

I already know from experimentation in Mathcad that an expression using this basic block produces a reasonable result (I suppose I should say bounded). The original expression is 2nd order and the associated response in s tends to 0 as s->0 and as s->INF. I managed to break down the original into a partial fraction sum so could treat it as 2 independent infinite sums of 1st order functions like the one shown. I didn't consider if/whether the 1st order expressions would diverge or not.

Perhaps I should post my original problem.

Last edited: Dec 17, 2013
5. ### blahdeblah

9
Here is my original problem:
$$FB(s)=\sum_{n=-\infty}^\infty \frac{a(k^ns)}{(k^ns)^2+a(k^ns)+1}$$
This is what I really want to obtain the closed form solution for.

6. ### jackmell

Entirely inadequate. Please restate the question precisely defining what a, k, and s are and not just "numbers" either.

7. ### blahdeblah

9
k is a real scalar > 1
a is a real scalar > 0
s is a imaginary scalar > 0

My first post asks if there is a closed form expression of the infinite sum given. If the answer is yes, then some guidance in the right direction to help to obtain it would be very helpful.

If a solution is indeed available then I think it follows that the expression in my last post (#6) can be solved.

Last edited: Dec 18, 2013
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