## Creating an Infinite Series with a Specific Property [closed]

Hello Mathematicians,

Recently I have been doing some work which has an interesting application to another problem as a consequence. However many searches for this latter problem have not returned a proper name for it, so I would be very grateful if you could help...

Problem: Specify an infinite sequence $a_n(\lambda)$, dependent on both $n$ and $\lambda$ in a non-trivial way, such that

$$\sum_n a_n(\lambda)^2=\rho\left(\sum_n a_n(\lambda)\right)^2,$$

where $\lambda$ may be altered to satisfy any finite value of the constant $\rho$.

I can solve this implicitly for $\rho\in\mathbb{R}$ under certain assumptions on the behaviour of $a_n(\lambda)$ and additionally can give methods for approximating $a_n(\lambda)$ to relatively high accuracy. The proof is a bit specialised and so I am mainly interested in more general work by others

Thank you for any help

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What are $\lambda$ and $\rho$ in this question? Parameters? Variables? I find the question hard to understand. – Yemon Choi Jun 1 2011 at 9:07
I don't really see why you need to use the letter $\lambda$ at all. It looks like you want a sequence of functions $a_n$ that depend only on $\rho$. – S. Carnahan Jun 1 2011 at 9:09
have you tried looking for finite sequences (with only 3, 4 or 5 terms) with $a_n(\lambda)$ given explicitely as, say, polynomial functions? – Olivier Bégassat Jun 1 2011 at 9:20
OK, so is $a_n(\lambda) = f(\lambda)b_n$ allowed, or not? If not, why not? Put more conditions on this $\lambda$ dependence! To be honest, this question looks like you're just making up random equations for its own sake; this can be great fun, but I don't see its relevance to mathematical research (unless you have a specific problem which needs this kind of construction, and even then it seems a bit too localised to me). – Zen Harper Jun 1 2011 at 10:29
I don't think this is a real question, since there is a large number of valid answers, which the OP does not like for some internal reason. Voting to close. – Igor Rivin Jun 1 2011 at 15:18

## closed as not a real question by Igor Rivin, Yemon Choi, David Loeffler, Simon Thomas, Gerry MyersonJun 1 2011 at 23:55

If $a_n=a_n(\lambda)=\lambda^n$ then $\sum a_n^2=\frac{1}{1-\lambda^2}$ while $(\sum a_n)^2=\frac{1}{(1-\lambda)^2}$ So $\lambda=\frac{1-\rho}{1+\rho}$ makes $\sum_n a_n(\lambda)^2=\rho\left(\sum_n a_n(\lambda)\right)^2$ for any $\rho \gt 0.$

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 FWIW, I think the OP wants $\rho$ to be specified and the identity to hold for all $\lambda$ (though given how ill-defined the original question is, I am largely guessing. – Yemon Choi Jun 1 2011 at 19:04