# Simplification of Summation

I am trying to simplify or analysis the convergence of the following equation. $c_k$ is between $0$ and $1$. Can someone please give an idea for that?

$$\frac{\Bigg(\sum_{k=1}^{K} \frac{2c_k}{a_k^2+b_k^2}\Bigg)^2}{\Bigg(\sum_{k=1}^{K} \frac{1}{a_k^2+b_k^2}\Bigg)^2}$$

• If two sequences $a_n, b_n$ converge to $A, B$, and say $B$ is nonzero, then $a_n/b_n$ converges to $A/B$. – Loki Clock Apr 3 '13 at 17:51
• It's not an equation, it is an expression. And it doesn't have a simplification. The square root is the weighted average of the value $2c_k$ with weights $(a_k^2+b_k^2)^{-1}$. So the resulting value is always between $0$ and $4$. – Thomas Andrews Apr 3 '13 at 17:56
• As for convergence, you'll have to know a lot more about these sequences, $a_i,b_i,c_i$ before you can say anything about convergence. I'd let $d_k=\frac{1}{a_k^2+b_k^2}$. Then you have $$\left(\frac{\sum 2c_kd_k}{\sum d_k}\right)^2$$ Without knowing more, you can't conclude much about the value other than it is between $0$ and $4$. – Thomas Andrews Apr 3 '13 at 18:00

The expressions does not simplify unless $K=1$ or there are some relations between $a_k, b_k, c_k$ not stated in the question.
Based on the information given, all we can say is that the ratio is between $0$ and $4$, which Thomas Andrews already observed:
$${\Bigg(\sum_{k=1}^{K} \frac{2c_k}{a_k^2+b_k^2}\Bigg)^2} \le 4 {\Bigg(\sum_{k=1}^{K} \frac{1}{a_k^2+b_k^2}\Bigg)^2}$$ because $c_k\le 1$.