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Any ideas on how to approximate and/or simplify this crazy-looking sum will be massively appreciated)

$$\frac{1}{\mu}\sum_{j=0}^{\frac{\lambda}{2}} \left(\frac{1}{2}+2\left(\frac{k}{n}\right)^2\right)^j\binom{\frac{\lambda}{2}}{j}\sum_{m=1}^{\mu} \left(\frac{m^2(m+2(\mu-m))^2}{\mu^4}\right)^j\left(1-\frac{m^2(m+2(\mu-m))^2}{\mu^4}\right)^{\frac{\lambda}{2}-j}$$

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1 Answer 1

up vote 5 down vote accepted

Summing over $j$ first, this is also $$\frac{1}{\mu}\sum_{m=1}^{\mu} \left(1-\left(\frac{1}{2}-2\left(\frac{k}{n}\right)^2\right)\frac{m^2(m+2(\mu-m))^2}{\mu^4}\right)^{\frac{\lambda}{2}} = \frac1{\mu^{1+2\lambda}}F(\alpha,\mu,\lambda/2), $$ for a value of $\alpha$ independent of $\mu$ and $\lambda$, with $$ F(u,r,i)=\sum_{m=1}^r\left(r^4-um^2(2r-m)^2\right)^i,\qquad \alpha=\frac{1}{2}-2\left(\frac{k}{n}\right)^2. $$ There probably does not exist any nice formula for $F(u,r,i)$ except for small values of $i$ or $r$, for example $F(u,1,i)=(1-u)^i$, or in special cases like $u=0$ since $F(0,r,i)=r^{4i+1}$ and one recovers the obvious value $1$ of the original sum when $n=2k$.

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thanks. Is there any proper name for $F(u,r,i)$? –  sigma.z.1980 Apr 24 '11 at 0:32
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