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Is there a closed form of this expression? 
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#1
Dec1713, 10:21 AM

P: 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. [tex] H(s)=\sum_{n=\infty}^\infty \frac{k^n}{k^n+a/s} [/tex] Thanks, D 


#2
Dec1713, 10:33 AM

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P: 4,500

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



#3
Dec1713, 10:33 AM

P: 9

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


#4
Dec1713, 10:37 AM

P: 9

Is there a closed form of this expression?
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. 


#5
Dec1713, 10:55 AM

P: 9

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


#6
Dec1813, 06:37 AM

P: 1,666




#7
Dec1813, 07:53 AM

P: 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. 


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