Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this sum, which probably doesn't exist in closed form.

$$\displaystyle S(\mu,\lambda,n,M,l)=\sum_{\alpha=1}^{\mu}\frac{1}{\alpha!}\Big(1-\frac{\alpha(2\mu-\alpha)(M-2l)}{2n\mu^2}\Big)^{\lambda} \leq \sum_{\alpha=1}^{\mu}\alpha^{\alpha+\frac{1}{2}} e^{-\alpha+\frac{\alpha \lambda(2\mu-\alpha)(M-2l)}{4n \mu^2}}$$ using Stirling number approximation and upper bound on $(1-\frac{k}{n})^n$.

Any ideas on how to find an asymptotic approximation for it would be massively appreciated. Please don't solve it for me, just give some hints/advice. Thanks

$\mu, \lambda,n,M$ are constants and $l$ is a variable. It will be used in the next summation of the form $S(\mu,\lambda,n,M,c)=\sum_{l=0}^{\frac{M}{2}-1}\frac{1}{c-S(\mu,\lambda,n,M,l)}$

share|cite|improve this question
I am unable to find $l$ anywhere in the sum. Typo? Maybe one of the terms should be $(M-2l)$? Given that the sum involves five parameters/variables, you need to specify more precisely what "asymptotic" means. – whuber Jun 3 '11 at 22:21
yes, I corrected accordingly. I'd like to get the answer in the form $f(\mu,\lambda,n,M,c)+O(g(\mu,\lambda,n,M))$ – sigma.z.1980 Jun 4 '11 at 0:26
Sometimes it's easier to get an approximation from some combinatorial/probabilistic argument - if the sum has such an interpretation? – leonbloy Jun 4 '11 at 3:02
$S_k$=Probablity of failure. Then $1-S_k$ is the probability of $k'th$ success/evolution, $\frac{1}{1-S_k}$ is the expected time until k'th success and summing over all k's I get the $S$, the expected time until a certain species has evolved $M$ times. Hope it helps. – sigma.z.1980 Jun 4 '11 at 3:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.