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? 



Determining the asymptotic behavior of a series
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf\_{n\to\infty}\ nt f\_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup\_{n\to\infty}\ nt f\_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f\_n(t) = \int\_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] $$
$$ = \int\_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int\_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f\_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int\_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int\_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf\_n(t)=\frac{-1}{\log\mu}+O\_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim\_{n\to\infty}n(1-t)^{n}=0$.


Based on the above argument I doubt that $\lim\_{n\to\infty}n f\_n(t)$ exists. More precisely, I don't think that
$$ \lim\_{n\to\infty} n\int\_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log\_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.


**EDIT:** It is easy to verify in an elementary fashion that $\lim\_{n\to\infty} nf\_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>1$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf\_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf\_n(t) \ll r^3 e^{-r} + r \ll r.$$
For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence
$$ n\sum\_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int\_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 e^{-r}, $$
while clearly
$$ n\sum\_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 e^{-r}+2r.$$
Hence $\lim\_{n\to\infty} nf\_n(t)$ does not exist when $r$ is sufficiently large.


