2
$\begingroup$

I have a series of the following form: \begin{align} \sum_{k=2}^\infty \left( 1 - e^{-ns^{k-1}} \right)^k \end{align} where $0<s<1$. I would like to compute an approximation of this series, for the regime where $n$ is large.

One natural thing to try would be to use the approximation $1-e^{-x} \approx x$, which is valid for small $x$. However, note that we are interested in large $n$, and the dominant terms of the series are those with small $k$, and thus $ns^{k-1}$ is relatively large for this approximation to be accurate. Also, even if it was an accurate approximation, I am not sure how to deal with the doubly exponential term that results from it.

Is there a better way to approximate this? I am particularly interested in a tight lower bound on this series, if it is practical to obtain one.

$\endgroup$
7
  • $\begingroup$ i would try euler mac laurin formula and approximate the resulting integral. $\endgroup$
    – tired
    Commented Dec 11, 2015 at 17:11
  • $\begingroup$ It seems to have an asymptotic like $$\sum_{k=2}^\infty \left( 1 - e^{-ns^{k-1}} \right)^k \approx \log \!\left(\frac{1}{s}\right) \Bigl[\log(n) - \log\log\log(n) - 1\Bigr],$$ though I haven't seen a way to obtain this rigorously yet. Numerically this also appears to be missing a term which is very small but may grow with $n$. $\endgroup$ Commented Dec 11, 2015 at 17:37
  • $\begingroup$ @tired I thought that too, but it seems tricky. I would definitely be interested to see an answer taking that approach. $\endgroup$ Commented Dec 11, 2015 at 17:40
  • $\begingroup$ @AntonioVargas , ok i will have a look at it when i have mathematica avaiable. $\endgroup$
    – tired
    Commented Dec 11, 2015 at 18:58
  • $\begingroup$ @AntonioVargas just out of interest, wich method did u use ? $\endgroup$
    – tired
    Commented Dec 11, 2015 at 20:09

1 Answer 1

0
$\begingroup$

I would rather keep the exponential form and do binomial expansion for say $k\le4$ (you may want to try a different number) and ignore terms for $5<k$. Then the form would be

\begin{align} \sum_{k=2}^{\infty} \left(1-e^{-ns^{k-1}}\right)^k&\approx 1-e^{-ns}+1-2e^{-ns^2}+e^{-2ns^2}+1-3e^{-ns^3}+3e^{-2ns^3}-e^{-3ns^3} \end{align} I'm not sure how much better this will be than your current approximation but as far as $n$ is a big number, this might help.

$\endgroup$
2
  • $\begingroup$ $0<s<1$ as stated in the question $\endgroup$
    – tired
    Commented Dec 11, 2015 at 15:31
  • $\begingroup$ @tired Yeah. Then we shouldn't use $(1-x)^k\approx 1-kx$. I'll change my answer. Thanks. $\endgroup$
    – Kay K.
    Commented Dec 11, 2015 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .