TITLE: Determining the asymptotic behavior of a series
QUESTION [8 upvotes]: I am trying to determine the behavior of the following series as $n\to\infty$. Let $0<\mu<1$ be fixed and for every positive integer $n\geq 1$, consider the function $f_n(t)$ of a real variable $t$ defined by the series $\sum_{k=0}^\infty\mu^k(1-\mu^kt)^n$. I want to determine how $f_n(t)$ behaves as $n\to\infty$ for $0<t<1$ (some kind of asymptotic formula).
Clearly $f_n(t)$ converges to $0$ for each $0<t<1$, but with what rate? And say, hypothetically, the rate is $O(1/n)$, then I would need to know at least what is $\lim_{n\to\infty}nf_n(t)$. I've tried several things for two weeks and I believe the rate of $O(1/n)$
is correct, but I can't find that limit. Any suggestions?

REPLY [2 votes]: Here is an elementary approach, which shows how  to find the nature of $nf_n(t)$ as $n\to\infty$. But I'm not going to bound error terms or such so this remains an outline until those details are filled in.
Write $a(k) = \mu^k(1-t\mu^k)^n$ and consider $k$ as a real variable.  $a(k)$ is easily seen to have a unique maximum at $k=k_0$ where $\mu^{-k_0}=1/((t(n+1))$. Now calculate
$$a(k_0) = \frac{(n/(n+1))^n}{t(n+1)} \sim \frac{e^{-1}}{t(n+1)}$$
$$a(k_0+u) = a(k_0) \mu^u\left(\frac{1-\mu^u/(n+1)}{1-1/(n+1)}\right)^n
 \sim a(k_0) \mu^u \exp(1-\mu^u)),$$
the limits being for $n\to\infty$ with $u$ not too wild.
So now we have (modulo checking of error terms),
$$nf_n(t) \sim \frac{1}{t} \sum_u \mu^u e^{-\mu^u},$$
where the sum is over $u\ge -k_0$ such that $k_0+u$ is an integer.
The restriction $u\ge -k_0$ should be far enough in the tail that it doesn't matter, so we have
$$nf_n(t) \sim \frac{1}{t} F_\mu(y),$$
where $y$ is the fractional part of $k_0$ and
$$F_\mu(y) = \sum_{j=-\infty}^\infty \mu^{y+j}e^{-\mu^{y+j}}.$$
(Doesn't this last sum have a name?  I'm sure I've seen it before.)